Trigonometry - Finding Trigonometric Roots

Which of the following is a solution to the following equation such that $0\leq x<2\pi?$

$\frac{11\pi}{6}$

$\frac{\pi}{6}$

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

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

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

Explanation

We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

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

We then determine the angles that satisfy each solution within one revolution.

The angles $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ satisfy the first, and $\frac{\pi}{2}$ satisfies the second. Only $\frac{11\pi}{6}$ is among our answer choices.

Solve the following equation for $0\leq x<2\pi$ .

$\sin 2x=\cos 2x$

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

$x=\frac{\pi}{8}; \frac{5\pi}{8}$

$x=\frac{\pi}{4}; \frac{5\pi}{4}$

$x=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}$

No solution exists

Explanation

The fastest way to solve this problem is to substitute a new variable. Let .

The equation now becomes:

$\sin u=\cos u$

So at what angles are the sine and cosine functions equal. This occurs at

$u=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}$

You may be wondering, "Why did you include

$\frac{9\pi}{4}; \frac{13\pi}{4}$ if they're not between and $2\pi$ ?"

The reason is because once we substitute back the original variable, we will have to divide by 2. This dividing by 2 will bring the last two answers within our range.

$2x=\frac{\pi}{4}; 2x=\frac{5\pi}{4}; 2x=\frac{9\pi}{4}; 2x=\frac{13\pi}{4}$

Dividing each answer by 2 gives us

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

Solve the equation for $0\leq x<2\pi$ .

$\cos 4x = \cos 2x$

$x=0; \frac{\pi}{3}; \frac{2\pi}{3}; \pi; \frac{4\pi}{3}; \frac{5\pi}{3};$

$x=0; \frac{\pi}{3}; \frac{2\pi}{3}$

$x=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};$

$x=0; \frac{2\pi}{3}; \frac{4\pi}{3}$

No solution exists

Explanation

We begin by substituting a new variable .

$\cos 2u=\cos u$ ; Use the double angle identity for $\cos 2u$ .

$2\cos ^2 u-1=\cos u$ ; subtract the $\cos u$ from both sides.

$2\cos ^2 u-\cos u-1=0$ ; This expression can be factored.

$(2\cos u+1)(\cos u-1)=0$ ; set each expression equal to 0.

$2\cos u+1=0$ or $\cos u -1=0$ ; solve each equation for $\cos u$

$\cos u = -\frac{1}{2}$ or $\cos u = 1$ ; Since we sustituted a new variable we can see that if $0\leq x<2\pi$ , then we must have $0\leq 2x < 4\pi$ . Since , that means $0\leq u<4\pi$ .

This is important information because it tells us that when we solve both equations for u, our answers can go all the way up to $4\pi$ not just $2\pi$ .

So we get

$2x = u=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};$ Divide everything by 2 to get our final solutions

$x=0; \frac{\pi}{3}; \frac{2\pi}{3}; \pi; \frac{4\pi}{3}; \frac{5\pi}{3};$

Solve the following equation for $0\leq x<2\pi$ .

$\sin\left(\frac{x}{2}\right)=\cos x -1$

$x=0, \frac{7\pi}{3}, \frac{11\pi}{3}$

$x=0, \frac{7\pi}{6}, \frac{11\pi}{6}$

$x=0, \frac{\pi}{6}, \frac{5\pi}{6}$

$x=0, \frac{\pi}{3}, \frac{5\pi}{3}$

Explanation

$\sin\left(\frac{x}{2}\right)=\cos x -1$ ; We start by substituting a new variable. Let $u=\frac{x}{2}$ .

$\sin u=\cos 2u-1$ ; Use the double angle identity for cosine

$\sin u = 1-2\sin ^2 u-1$ ; the 1's cancel, so add $2\sin ^2u$ to both sides

$2\sin ^2u+\sin u=0$ ; factor out a $\sin u$ from both terms.

$\sin u(2\sin u+1)=0$ ; set each expression equal to 0.

$\sin u =0$ or $2\sin u +1=0$ ; solve the second equation for sin u.

$\sin u = 0$ or $\sin u = -\frac{1}{2}$ ; take the inverse sine to solve for u (use a unit circle diagram or a calculator)

$\frac{x}{2}=u=0, \frac{7\pi}{6}, \frac{11\pi}{6}$ ; multiply everything by 2 to solve for x.

$x=0, \frac{7\pi}{3}, \frac{11\pi}{3}$ ; Notice that the last two solutions are not within our range $0\leq x<2\pi$ . So the only solution is .

Solve the following equation for $0\leq x<2\pi$ .

$4\cos ^2x=3$

$x=\frac{\pi}{6}; \frac{5\pi}{6}; \frac{7\pi}{6}; \frac{11\pi}{6}$

$x=\frac{\pi}{6}; \frac{11\pi}{6}$

$x=\frac{5\pi}{6}; \frac{7\pi}{6}$

$x=\frac{\pi}{3}; \frac{2\pi}{3}; \frac{4\pi}{3}; \frac{5\pi}{3};$

No solution exists

Explanation

$4\cos ^2x=3$ ; First divide both sides of the equation by 4

$\cos ^2x=\frac{3}{4}$ ; Next take the square root on both sides. Be careful. Remember that when YOU take a square root to solve an equation, the answer could be positive or negative. (If the square root was already a part of the equation, it usually only requires the positive square root. For example, the solutions to are 2 and -2, but if we plug in 4 into the function $f(x)=\sqrt{x}$ the answer is only 2.) So,

$\cos x = \pm \frac{\sqrt{3}}{2}$ ; we can separate this into two equations

$\cos x = \frac{\sqrt{3}}{2}$ and $\cos x =- \frac{\sqrt{3}}{2}$ ; we get

$x=\frac{\pi}{6}; \frac{11\pi}{6}$ and $x=\frac{5\pi}{6}; \frac{7\pi}{6}$

Solve the equation for $0\leq x<2\pi$ .

$3\tan^2x=1$

$x=\frac{\pi}{6}; \frac{5\pi}{6}; \frac{7\pi}{6}; \frac{11\pi}{6}$

$x=\frac{\pi}{6}; \frac{7\pi}{6}$

$x=\frac{5\pi}{6}; \frac{11\pi}{6}$

$x=\frac{\pi}{3}; \frac{2\pi}{3}; \frac{4\pi}{3}; \frac{5\pi}{3}$

$x=\frac{\pi}{3}; \frac{4\pi}{3}$

Explanation

$3\tan^2x=1$ ; Divide both sides by 3

$\tan^2x=\frac{1}{3}$ ; Take the square root on both sides. Just as the previous question, when you take a square root the answer could be positive or negative.

$\tan x=\pm \frac{1}{\sqrt{3}}$ ; This can be written as two separate equations

$\tan x=\frac{1}{\sqrt{3}}$ and $\tan x=-\frac{1}{\sqrt{3}}$ ; Take the inverse tangent

$x=\frac{\pi}{6}; \frac{7\pi}{6}$ and $x=\frac{5\pi}{6}; \frac{11\pi}{6}$

Solve the following equation for $0\leq x<2\pi$ .

$\tan ^2x+2\tan x+1=0$

$x=\frac{3\pi}{4}; \frac{7\pi}{4}$

$x=\frac{\pi}{4}; \frac{3\pi}{4}; \frac{5\pi}{4}; \frac{7\pi}{4}$

$x=\frac{\pi}{4}; \frac{5\pi}{4}$

$x=\frac{\pi}{4}; \frac{3\pi}{4}$

$x=\frac{5\pi}{4}; \frac{7\pi}{4}$

Explanation

$\tan ^2x+2\tan x+1=0$ ; The expression is similar to a quadratic expression and can be factored.

$(\tan x+1)(\tan x+1)=0$ ; set both expressions equal to 0. Since they are the same, the solutions will repeat, so I will only write it once.

$\tan x+1=0$

$\tan x=-1$ ; take the inverse tangent on both sides

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

$x=\frac{3\pi}{4}; \frac{7\pi}{4}$

Solve the equation below for greater than or equal to and strictly less than .

and

only

and

Explanation

Recall the values of for which . If it helps, think of sine as the values on the unit circle. Thus, the acceptable values of would be 0, 180, 360, 540 etc.. However, in our scenario .

Thus we have and .

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

and .

Solve the following equation. Find all solutions such that $0\leq x <2\pi$ .

$2\sin x = 1$

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

$x=\frac{\pi}{3}; \frac{5\pi}{3}$

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

$x=\frac{\pi}{6}; \frac{5\pi}{6}; \frac{7\pi}{6}; \frac{11\pi}{6}$

No Solution

Explanation

$2\sin x=1$ ; Divide both sides by 2 to get

$\sin x=\frac{1}{2}$ ; take the inverse sine on both sides

$\sin^{-1}(\sin x)=\sin ^{-1}\bigg(\frac{1}{2}\bigg)$ ; the left side reduces to x, so

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

At this point, either use a unit circle diagram or a calculator to find the value.

Keep in mind that the problem asks for all solutions between and $2\pi$ .

If you use a calculator, you will only get $\frac{\pi}{6}$ as an answer.

So we need to find another angle that satisfies the equation $\sin x=\frac{1}{2}$ .
Problem_1_set_1

Solve the following equation. Find all solutions such that $0\leq x<2\pi$ .

$\sin 2x=4\cos x$

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

No solution exists

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

$x=0;\pi$

Explanation

$\sin 2x=4\cos x$ ; First use the double angle identity for $\sin 2x$ .

$2\sin x \cos x = 4\cos x$ ; divide both sides by 2

$\sin x \cos x=2\cos x$ ; subtract the $2\cos x$ from both sides

$\sin x \cos x-2\cos x=0$ ; factor out the $\cos x$

$\cos x(\sin x-2)=0$ ; Now we have the product of two expressions is 0. This can only happen if one (or both) expressions are equal to 0. So let each expression equal 0.