Solve Trigonometric Functions with Inverse Functions: CCSS.Math.Content.HSF-TF.B.7 - Common Core: High School - Functions

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Solve the following for $\theta$

$2\tan(\theta)+4=6$

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This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)+4=6$

First subtract four from each side.

$2\tan(\theta)+4{\color{Red} -4}=6{\color{Red} -4}$

$2\tan(\theta)=2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}$=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=1$

$\tan\left($\frac{5\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\frac{\pi}{4}$,$\frac{5\pi}{4}$ .

Solve the following for $\theta$ .

$2\tan(\theta)=2$

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This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2$

Divide each side by two.

$\frac{2\tan(\theta)}{2}$=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=1$

$\tan\left($\frac{5\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{4}$,$\frac{5\pi}{4}$$

Solve for $\theta$ .

$\tan(\theta)+4=$\sqrt{3}$+4$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)+4=$\sqrt{3}$+4$

First subtract four from both sides.

$\\tan(\theta)+4{\color{Red} -4}=$\sqrt{3}$+4{\color{Red} -4} \\tan(\theta)=$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$($\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{3}$ \right )=\frac{\frac{\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot $\frac{2}{1}$=$\sqrt{3}$$

$\tan\left($\frac{4\pi}{3}$ \right )=\frac{\frac{-\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{-2}{1}$=$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3}$,$\frac{4\pi}{3}$$

Solve for $\theta$ .

$2\tan(\theta)=2$\sqrt{3}$$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2$\sqrt{3}$$

Divide by two on both sides.

$$\frac{2\tan(\theta)}{2}$=\frac{2\sqrt{3}$}{2}$

$\tan(\theta)=$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$($\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{3}$ \right)=\frac{\frac{\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot $\frac{2}{1}$=$\sqrt{3}$$

$\tan\left($\frac{4\pi}{3}$ \right)=\frac{\frac{-\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{-2}{1}$=$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3}$,$\frac{4\pi}{3}$$

Solve for $\theta$ .

$3\tan(\theta)=\frac{-3}{\sqrt{3}$}$

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This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{-3}{\sqrt{3}$}$

Divide by three on both sides.

$$\frac{3\tan(\theta)}{3}$=\frac{-3}{3\sqrt{3}$}$

$\tan(\theta)=-$\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left(-$\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{5\pi}{6}$ \right )=\frac{\frac{1}{2}$}{-$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{11\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{5\pi}{6}$,$\frac{11\pi}{6}$$

Solve for $\theta$ .

$3\tan(\theta)=\frac{3}{\sqrt{3}$}$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{3}{\sqrt{3}$}$

Divide by three on both sides.

$$\frac{3\tan(\theta)}{3}$=\frac{3}{3\sqrt{3}$}$

$\tan(\theta)=\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left($\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{6}$ \right)=\frac{\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{7\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{-\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{6}$,$\frac{7\pi}{6}$$

Solve for $\theta$ .

$\tan(\theta)-1=\frac{1}{\sqrt{3}$}-1$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)-1=\frac{1}{\sqrt{3}$}-1$

Add one to both sides.

$\tan(\theta)-1{\color{Red} +1}=\frac{1}{\sqrt{3}$}-1{\color{Red} +1}$

$\tan(\theta)=\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left($\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{6}$ \right)=\frac{\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{7\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{-\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{6}$,$\frac{7\pi}{6}$$

Solve the following for $\theta$ .

$2\tan(\theta)-4=-6$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)-4=-6$

First add four from each side.

$2\tan(\theta)-4{\color{Red} +4}=-6{\color{Red} +4}$

$2\tan(\theta)=-2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}$=\frac{-2}{2}$

$\tan(\theta)=-1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(-1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{3\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=-1$

$\tan\left($\frac{7\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=-1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{3\pi}{4}$,$\frac{7\pi}{4}$$

Solve for $\theta$ .

$\tan(\theta)+1=-$\sqrt{3}$+1$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)+1=-$\sqrt{3}$+1$

Subtract one from each side.

$\tan(\theta)+1{\color{Red} -1}=-$\sqrt{3}$+1{\color{Red} -1}$

$\tan(\theta)=-$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(-$\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{2\pi}{3}$ \right )=\frac{\frac{\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot -$\frac{2}{1}$=-$\sqrt{3}$$

$\tan\left($\frac{5\pi}{3}$ \right )=\frac{\frac{-\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{2}{1}$=-$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{2\pi}{3}$,$\frac{5\pi}{3}$$

Solve for $\theta$ .

$2\tan(\theta)=-2$\sqrt{3}$$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)=-2$\sqrt{3}$$

Divide by two on each side.

$$\frac{2\tan(\theta)}{2}$=\frac{-2\sqrt{3}$}{2}$

$\tan(\theta)=-$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(-$\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{2\pi}{3}$ \right)=\frac{\frac{\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot -$\frac{2}{1}$=-$\sqrt{3}$$

$\tan\left($\frac{5\pi}{3}$ \right)=\frac{\frac{-\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{2}{1}$=-$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{2\pi}{3}$,$\frac{5\pi}{3}$$

Solve for $\theta$ .

$\tan(\theta)+1=1-$\frac{1}{\sqrt{3}$}$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)+1=1-$\frac{1}{\sqrt{3}$}$

Subtract one from each side.

$\tan(\theta)+1{\color{Red} -1}=1-$\frac{1}{\sqrt{3}$}{\color{Red} -1}$

$\tan(\theta)=-$\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left(-$\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{5\pi}{6}$ \right )=\frac{\frac{1}{2}$}{$\frac{-\sqrt{3}$}{2}}=\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{11\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{5\pi}{6}$,$\frac{11\pi}{6}$$

Solve for $\theta$ .

$-\tan(\theta)+2=3$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$-\tan(\theta)+2=3$

Subtract two from both sides.

$-\tan(\theta)+2{\color{Red} -2}=3{\color{Red} -2}$

$-\tan(\theta)=1$

Now divide by negative one on both sides.

$\tan(\theta)=-1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(-1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{3\pi}{4}$ \right )=\frac{\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=-1$

$\tan\left($\frac{7\pi}{4}$ \right )=\frac{-\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=-1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{3\pi}{4}$,$\frac{7\pi}{4}$$

Solve the following for $\theta$

$2\tan(\theta)+4=6$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)+4=6$

First subtract four from each side.

$2\tan(\theta)+4{\color{Red} -4}=6{\color{Red} -4}$

$2\tan(\theta)=2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}$=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=1$

$\tan\left($\frac{5\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\frac{\pi}{4}$,$\frac{5\pi}{4}$ .

Solve the following for $\theta$ .

$2\tan(\theta)=2$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2$

Divide each side by two.

$\frac{2\tan(\theta)}{2}$=\frac{2}{2}$

$\tan(\theta)=1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=1$

$\tan\left($\frac{5\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{4}$,$\frac{5\pi}{4}$$

Solve for $\theta$ .

$\tan(\theta)+4=$\sqrt{3}$+4$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)+4=$\sqrt{3}$+4$

First subtract four from both sides.

$\\tan(\theta)+4{\color{Red} -4}=$\sqrt{3}$+4{\color{Red} -4} \\tan(\theta)=$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$($\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{3}$ \right )=\frac{\frac{\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot $\frac{2}{1}$=$\sqrt{3}$$

$\tan\left($\frac{4\pi}{3}$ \right )=\frac{\frac{-\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{-2}{1}$=$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3}$,$\frac{4\pi}{3}$$

Solve for $\theta$ .

$2\tan(\theta)=2$\sqrt{3}$$

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This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)=2$\sqrt{3}$$

Divide by two on both sides.

$$\frac{2\tan(\theta)}{2}$=\frac{2\sqrt{3}$}{2}$

$\tan(\theta)=$\sqrt{3}$$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$($\sqrt{3}$)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{3}$ \right)=\frac{\frac{\sqrt{3}$}{2}}{$\frac{1}{2}$}=\frac{\sqrt{3}$}{2}\cdot $\frac{2}{1}$=$\sqrt{3}$$

$\tan\left($\frac{4\pi}{3}$ \right)=\frac{\frac{-\sqrt{3}$}{2}}{-$\frac{1}{2}$}=\frac{-\sqrt{3}$}{2}\cdot $\frac{-2}{1}$=$\sqrt{3}$$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{3}$,$\frac{4\pi}{3}$$

Solve for $\theta$ .

$3\tan(\theta)=\frac{-3}{\sqrt{3}$}$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{-3}{\sqrt{3}$}$

Divide by three on both sides.

$$\frac{3\tan(\theta)}{3}$=\frac{-3}{3\sqrt{3}$}$

$\tan(\theta)=-$\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left(-$\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{5\pi}{6}$ \right )=\frac{\frac{1}{2}$}{-$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{11\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=-$\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{5\pi}{6}$,$\frac{11\pi}{6}$$

Solve for $\theta$ .

$3\tan(\theta)=\frac{3}{\sqrt{3}$}$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$3\tan(\theta)=\frac{3}{\sqrt{3}$}$

Divide by three on both sides.

$$\frac{3\tan(\theta)}{3}$=\frac{3}{3\sqrt{3}$}$

$\tan(\theta)=\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left($\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{6}$ \right)=\frac{\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{7\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{-\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{6}$,$\frac{7\pi}{6}$$

Solve for $\theta$ .

$\tan(\theta)-1=\frac{1}{\sqrt{3}$}-1$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$\tan(\theta)-1=\frac{1}{\sqrt{3}$}-1$

Add one to both sides.

$\tan(\theta)-1{\color{Red} +1}=\frac{1}{\sqrt{3}$}-1{\color{Red} +1}$

$\tan(\theta)=\frac{1}{\sqrt{3}$}$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$\left($\frac{1}{\sqrt{3}$}\right)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{\pi}{6}$ \right)=\frac{\frac{1}{2}$}{$\frac{\sqrt{3}$}{2}}=\frac{1}{2}$\cdot $\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

$\tan\left($\frac{7\pi}{6}$ \right )=\frac{-\frac{1}{2}$}{$\frac{-\sqrt{3}$}{2}}=-$\frac{1}{2}$\cdot -$\frac{2}{\sqrt{3}$}=\frac{1}{\sqrt{3}$}$

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{\pi}{6}$,$\frac{7\pi}{6}$$

Solve the following for $\theta$ .

$2\tan(\theta)-4=-6$

Tap to see back →

This question is testing ones ability to understand and identify inverses of trigonometric functions in equations as they relate to the unit circle.

For the purpose of Common Core Standards, "Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context" concept (CCSS.MATH.CONTENT.HSF-TF.B.7). It is important to note that this standard is not directly tested on but is used for building a deeper understanding on trigonometric functions.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function.

$2\tan(\theta)-4=-6$

First add four from each side.

$2\tan(\theta)-4{\color{Red} +4}=-6{\color{Red} +4}$

$2\tan(\theta)=-2$

Divide both sides by two.

$\frac{2\tan(\theta)}{2}$=\frac{-2}{2}$

$\tan(\theta)=-1$

Step 2: To isolate theta, perform the inverse trigonometric operation.

$\ $\tan^{-1}$$(\tan(\theta))=\tan^{-1}$(-1)$

Step 3: Use the unit circle to solve for theta.

Recall that

$\tan\theta=\frac{\sin\theta}{\cos\theta}$$

Therefore, to have tangent equalling one, both sine and cosine must be equal to one another.

Therefore,

$\tan\left($\frac{3\pi}{4}$ \right)=\frac{\frac{1}{\sqrt2}$}{-$\frac{1}{\sqrt2}$}=\frac{1}{\sqrt2}$\cdot -$\frac{\sqrt2}{1}$=-1$

$\tan\left($\frac{7\pi}{4}$ \right)=\frac{-\frac{1}{\sqrt2}$}{$\frac{1}{\sqrt2}$}=-$\frac{1}{\sqrt2}$\cdot $\frac{\sqrt2}{1}$=-1$ .

Step 4: Answer the question.

Solving for $\theta$ results in two possible values,

$\theta=\frac{3\pi}{4}$,$\frac{7\pi}{4}$$