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Trigonometry : Finding Trigonometric Roots

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Example Questions

Example Question #219 : Trigonometry

Solve for $\theta$ using a trigonometric identity: $\sin \theta + \cos 2 \theta = -\frac{ 7} {8 }$

Possible Answers:

48.59 ^ o , 311. 41 ^ o

311. 41 ^ o , 221.41 ^ o

311.41 ^ o , 228. 59 ^ o

131.41 ^ o , 311. 41 ^ o

-48.59 ^ o , 48. 5 9 ^ o

Correct answer:

311.41 ^ o , 228. 59 ^ o

Explanation:

First, using the trigonometric identity $\cos 2 x = 1 - 2 \sin ^ 2 2x$ , re-write the expression so it is entirely in terms of $\sin \theta$ :

$\sin \theta + 1 - 2 \sin ^2 \theta = -\frac{ 7}{8 }$ if we add $\frac{ 7} {8}$ to both sides and re-order the terms, we can see that this is a quadratic equation in terms of sine:

$-2 \sin ^ 2 \theta + \sin \theta + 1 + \frac{ 7}{8 } = 0$

$1 + \frac{ 7}{8 } = \frac{ 15}{8 }$ so we can use the quadratic formula to solve this way:

$\sin \theta = \frac{ -1 \pm \sqrt{ 1 - 4(-2)(\frac{15}{8 })}}{2(-2)} = \frac {-1 \pm \sqrt{1 + 15}}{-4 } = \frac{-1 \pm 4 }{-4}$

This gives us two potential answers:

$\sin \theta = \frac{ -1 -4 }{-4 } = \frac{ -5 }{-4 } = 1.25$ this is outside of the range of sine, so it won't work.

$\sin \theta = \frac{ -1 + 4 }{-4 } = \frac{ 3}{-4 } = -0.75$

Taking the inverse on a calculator yields -48.59 , which we can add to 360 to give 311.41^o .

That's just one answer, however. To get the other answer, we need to know the other angle 48.59 ^o below the x-axis, so we can add 180^o + 48.59 ^ o = 228.59 ^ o .

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Example Question #220 : Trigonometry

Find the roots of the equation $y = \sin(2 \theta ) \cos \theta + \cos \theta - \sin (2 \theta ) - 1$

Possible Answers:

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

$\frac{ 3 \pi }{4} , 2 \pi$

$\frac{ 3 \pi }{2} , \pi$

$\frac {\pi }{4} , 2 \pi$

$\frac{ 3 \pi }{2} , 0$

Correct answer:

$\frac{ 3 \pi }{4} , 2 \pi$

Explanation:

We can factor this expression by grouping.

$y = \sin(2 \theta ) \cos \theta + \cos \theta - \sin (2 \theta ) - 1$ re-arrange the original expression so that the terms with sine are next to each other

$y = \sin (2 \theta ) \cos \theta - \sin (2 \theta ) + \cos \theta - 1$ factor our the $\sin (2 \theta)$

$y = \sin (2 \theta ) [ \cos \theta - 1 ] + \cos \theta - 1$ this factors to

$y = (\cos \theta - 1 ) (\sin ( 2 \theta ) + 1 )$

Since we are finding the roots, we should set both factors equal to 0:

$\cos \theta - 1 = 0$ add 1 to both sides

$\cos \theta = 1$

$\theta = cos ^ {-1} (1) = 2 \pi$

$\sin ( 2 \theta ) + 1 = 0$ subtract 1 from both sides

$\sin ( 2 \theta ) = - 1$

$2 \theta = \sin ^ { -1 } (-1)$

$2 \theta = \frac{ 3 \pi }{2}$ divide by 2

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

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Example Question #21 : Finding Trigonometric Roots

Find the roots of the equation $y = \frac{ \sqrt 3 }{3} - \frac{ 2}{3} \sin (\frac{ \theta }{2})$

Possible Answers:

no solution

$\frac{ 2 \pi }{3} , \frac{ 4 \pi }{3}$

$\frac{ \pi }{3} , \frac{ 2 \pi }{3}$

$\frac{ \pi }{3} , \frac{ 11 \pi }{3}$

$\frac{ \pi }{6} , \frac{ \pi }{3}$

Correct answer:

$\frac{ 2 \pi }{3} , \frac{ 4 \pi }{3}$

Explanation:

There are many ways to solve this problem, and here is one.

First, re-write the equation so that it is set equal to zero, since we're finding the roots, and so that it is all in one fraction. That is easy to do since both terms already have the common denominator of 3:

$\frac{ \sqrt3 - 2 \sin (\frac{\theta }{2} ) } {3 } = 0$ multiply both sides by 3

$\sqrt3 - 2 \sin (\frac{ \theta }{ 2 }) = 0$ add the sine term to both sides

$\sqrt3 = 2 \sin (\frac{ \theta }{2 })$ divide both sides by 2

$\frac{ \sqrt3 }{2} = \sin (\frac{ \theta }{2} )$

$\sin ^ {-1 } (\frac{\sqrt3}{2} ) = \frac{ \theta }{2}$

$\frac{ \theta }{2} = \frac{ \pi }{3} , \frac{ 2 \pi }{3}$ multiply by 2

$\theta = \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3}$

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Example Question #34 : Solving Trigonometric Equations

Find the roots of the equation $2 \cos ^2 (\frac{ 2 }{3} \theta ) \sin \theta - \sin \theta + 10 \cos ^2 (\frac{2}{3} \theta ) - 5 = y$

Possible Answers:

$\frac{ \pi }{6} , \frac{ 7 \pi }{6} , \frac{ \pi }{2} , \frac{ 5 \pi }{6}$

$\frac{ 3 \pi }{8 } , \frac{ 21 \pi }{8 }$

$\frac{ \pi }{4} , \frac{ 7 \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}$

$\frac{ 3 \pi }{8 } , \frac{ 5 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

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

Correct answer:

$\frac{ 3 \pi }{8 } , \frac{ 5 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

Explanation:

Factor the equation by grouping: the first 2 terms have $\sin \theta$ in common, and the second 2 have 5 in common:

$y = \sin \theta [ 2 \cos ^2 (\frac{ 2}{3 } \theta ) - 1 ] + 5 [ 2 \cos ^2 (\frac{2}{3} \theta ) - 1 ]$

$y = (2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 ) ( \sin \theta + 5 )$

Since we are finding the roots, we can set each factor equal to 0:

$\sin \theta + 5 = 0$ subtract 5 from both sides

$\sin \theta = -5$ this is outside of the range for sine

$2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 = 0$ add 1 to both sides

$2 \cos ^ 2 ( \frac{ 2}{3 } \theta ) = 1$ divide both sides by 2

$\cos ^2 ( \frac{ 2}{3 }\theta ) = \frac{ 1}{2}$ take the square root of both sides

$\cos (\frac{ 2}{3} \theta ) = \pm \frac{ 1 }{\sqrt2}$

$\frac{2}{3} \theta = \cos ^ {-1} ( \pm \frac{ 1}{ \sqrt2})$

$\frac{ 2}{3} \theta = \frac{ \pi }{4} , \frac{ 7 \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}$ multiply by $\frac{ 3}{2 }$

$\theta = \frac{ 3 \pi }{8 } , \frac{ 21 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

That second one is greater than $2\pi$ , so we want to figure out which angle between 0 and $2 \pi$ is coterminal with it. We can figure this out by subtracting $2 \pi$ , or in this case $\frac{ 16 \pi }{ 8 }$ to get $\frac{ 21 \pi }{ 8 } - \frac{ 16 \pi }{ 8 } = \frac{ 5 \pi } {8 }$

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Example Question #22 : Finding Trigonometric Roots

Find the roots of the equation $y= 2 \sin \theta \cos (4 \theta ) - \sqrt3 \cos (4 \theta ) + 2 \sin \theta - \sqrt3$

Possible Answers:

$\pi , 2 \pi$

$\frac{ 4 \pi }{3} , \frac{ 5 \pi }{3} , \frac{ \pi }{4}$

$\frac{ \pi }{3} , \pi$

$\frac{ \pi }{3} , \frac{\pi }{4} , \frac{ 2 \pi }{3}$

$\frac{4\pi }{3} , \frac{ 5 \pi }{3} , 4 \pi$

Correct answer:

$\frac{ \pi }{3} , \frac{\pi }{4} , \frac{ 2 \pi }{3}$

Explanation:

Factor by grouping. The first 2 terms have $\cos(4 \theta)$ in common.

$y = \cos (4\theta ) [ 2 \sin \theta - \sqrt3 ] + 1[2 \sin \theta - \sqrt3]$

$y = ( \cos (4 \theta ) + 1 ) ( 2 \sin \theta - \sqrt3 )$

We are finding the roots, so set each factor equal to 0:

$\cos (4 \theta ) + 1 = 0$ subtract 1 from both sides

$\cos (4 \theta ) = -1$

$4 \theta = \cos ^ {-1 } (-1)$

$4 \theta = \pi$ divide by 4

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

$2 \sin \theta - \sqrt3 = 0$ add $\sqrt3$ to both sides

$2 \sin \theta = \sqrt 3$ divide both sides by 2

$\sin \theta = \frac{ \sqrt3}{2 }$

$\theta = \sin ^ { -1 } (\frac{ \sqrt3}{2} )$

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

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Example Question #23 : Finding Trigonometric Roots

Find the roots of the equation $y = 2 \sin x \cos x - \sin x - 2 \cos x + 1$

Possible Answers:

$x = \frac{\pi }{3} , \frac{2 \pi }{3} , \pi$

$x = \frac{\pi }{6} , \frac{\pi }{2}$

$x = \frac{\pi }{6} , \frac{5 \pi }{6} , \frac{\pi}{2}$

$x = \frac{\pi }{3} , \frac{2 \pi }{3} , \frac{3 \pi }{2}$

$x = \frac{ \pi }{6} , \frac{5 \pi }{6} , \pi$

Correct answer:

$x = \frac{\pi }{6} , \frac{5 \pi }{6} , \frac{\pi}{2}$

Explanation:

Factor the equation:

$y = \sin x (2 \cos x - 1 ) - 1 (2 \cos x - 1 )$ $y = (\sin x - 1 )(2 \cos x - 1 )$

The roots occur when each factor equals 0:

$\sin x - 1 = 0$

$\sin x = 1$

$x = \sin ^{-1}(1) = \frac{\pi }{2}$

$2 \cos x - 1 = 0$

$2 \cos x = 1$

$\cos x = \frac{1}{2}$

$x = \cos ^{-1}(\frac{1}{2}) = \frac{\pi }{6} , \frac{5 \pi }{6}$

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Example Question #24 : Finding Trigonometric Roots

Find all roots of this equation, in radians.

$\tan^2 x=3/4$

Possible Answers:

x=0.64, x=-0.64

x=0.64

x=-2.28

x=0.71, x=-0.71

x=1.04, x=-1.04

Correct answer:

x=0.71, x=-0.71

Explanation:

Take the square root of both sides:

$\tan^2 x=3/4\rightarrow\tan x=\pm\sqrt{3}/2$

This gives us two roots,

$\tan x=\sqrt{3}/2$ and $\tan x=-\sqrt{3}/2$

Solving for x:

$x=\arctan (\sqrt{3}/2)$ or $x=\arctan (-\sqrt{3}/2)$ ,

which evaluate to x=0.71 and x=-0.71

Now we check to make sure both answers work. Make sure your calculator is in radians mode!

$\tan^2 (0.71)=0.74\approx 3/4$

$\tan^2 (-0.71)=0.74\approx 3/4$

Our answers are x=0.71, x=-0.71

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Example Question #25 : Finding Trigonometric Roots

If $\small \tan x * \cos x = 1$ What is in degrees?

Possible Answers:

Correct answer:

Explanation:

if $\small \tan x$ is $\small \frac{\sin x }{\cos x }$

$\frac{\sin(x)}{\cos(x)}\cdot \cos(x)=1$

$\sin(x)=1$

$x=\sin^{-1}(1)=90$

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Trigonometry : Finding Trigonometric Roots

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #219 : Trigonometry

Example Question #220 : Trigonometry

Example Question #21 : Finding Trigonometric Roots

Example Question #34 : Solving Trigonometric Equations

Example Question #22 : Finding Trigonometric Roots

Example Question #23 : Finding Trigonometric Roots

Example Question #24 : Finding Trigonometric Roots

Example Question #25 : Finding Trigonometric Roots

All Trigonometry Resources

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