# 3.4.4: Solving Equations with Double-Angle Identities

• • Contributed by CK12
• CK12

Solve sine, cosine, and tangent of angles multiplied or divided by 2.

Trig Riddle: I am an angle x such that $$0\leq x<2\pi$$. I satisfy the equation $$\sin 2x−\sin x=0$$. What angle am I?

### Solve Trigonometric Equations

We can use the half and double angle formulas to solve trigonometric equations.

Let's solve the following trigonometric equations.

1. Solve $$\tan 2x+\tan x=0$$ when $$0\leq x<2\pi$$.

Change $$\tan 2x$$ and simplify.

\begin{aligned} \tan 2x+\tan x &=0\\ \dfrac{2\tan x}{1−\tan ^2x}+\tan x &=0\\ 2\tan x+\tan x(1−\tan ^2x) &=0\rightarrow \text{Multiply everything by } 1−\tan ^2x \text{ to eliminate denominator. }\\ 2\tan x+\tan x−\tan ^3x&=0 \\ 3\tan x−\tan ^3x&=0 \\ \tan x(3−\tan ^2x)&=0 \end{aligned}

Set each factor equal to zero and solve.

\begin{aligned} & & 3−\tan ^2 x&=0 \\ & & −\tan ^2x&=−3 \\ \tan x&=0 &\text{ and } \qquad \tan ^2x &=3 \\ x&=0 \text{ and } \pi & \tan x&=\pm \sqrt{3} \\ & & x&=\dfrac{\pi}{3},\; \dfrac{2\pi}{3},\; \dfrac{4\pi}{3},\; \dfrac{5 \pi}{3} \end{aligned}

1. Solve $$2\cos \dfrac{x}{2}+1=0$$ when $$0\leq x<2\pi$$.

In this case, you do not have to use the half-angle formula. Solve for $$\dfrac{x}{2}$$.

\begin{aligned}2\cos \dfrac{x}{2}+1=0 \\ 2\cos \dfrac{x}{2}=−1 \\ \cos \dfrac{x}{2}=−\dfrac{1}{2} \end{aligned}

Now, let’s find $$\cos a=−\dfrac{1}{2}$$ and then solve for x by dividing by 2.

\begin{aligned} \dfrac{x}{2}&=\dfrac{2 \pi}{3},\dfrac{4 \pi}{3} \\ &=\dfrac{4 \pi}{3},\; \dfrac{8 \pi}{3} \end{aligned}

Now, the second solution is not in our range, so the only solution is $$x=\dfrac{4 \pi}{3}$$.

1. Solve $$4\sin x\cos x=\sqrt{3}$$ for $$0\leq x<2\pi$$.

Pull a 2 out of the left-hand side and use the $$\sin 2x$$ formula.

\begin{aligned} 4\sin x\cos x&=\sqrt{3} \\ 2\cdot 2\sin x\cos x&=\sqrt{3} \\ 2\cdot \sin 2x&=\sqrt{3} \\ \sin 2x&=\dfrac{\sqrt{3} }{2} \\ 2x &=\dfrac{\pi}{3},\; \dfrac{5 \pi}{3},\; \dfrac{7 \pi}{3},\; \dfrac{11 \pi}{3} \\ x&=\dfrac{\pi}{6},\; \dfrac{5 \pi}{6},\; \dfrac{7 \pi}{6},\; \dfrac{11 \pi}{6} \end{aligned}

Example $$\PageIndex{1}$$

Earlier, you were asked to find the angle x, where $$0\leq x<2\pi$$, such that $$x$$ satisfies the equation $$\sin 2x−\sin x=0$$.

Solution

Use the double angle formula and simplify.

\begin{aligned} \sin 2x−\sin x&=0 \\ 2\sin x\cos x−\sin x&=0 \\ \sin x(2\cos x−1)&=0 \\ \sin x=0 \text{ OR } \cos x&=\dfrac{1}{2} \end{aligned}

Under the constraint $$0\leq x<2\pi$$, $$\sin x=0$$ when $$x=0$$ or when $$x=\pi$$. Under this same constraint, $$\cos x=\dfrac{1}{2}$$ when $$x=\dfrac{\pi }{3}$$ or when $$x=\dfrac{5 \pi}{3}$$.

Example $$\PageIndex{2}$$

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

$$\sin \dfrac{x}{2}=−1$$

Solution

\begin{aligned} \sin \dfrac{x}{2}&=−1 \\ \dfrac{x}{2} &=\dfrac{3\pi }{2}\\ x&=3\pi \end{aligned}

From this we can see that there are no solutions within our interval.

Example $$\PageIndex{3}$$

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

$$\cos 2x−\cos x=0$$

Solution

\begin{aligned} \cos 2x−\cos x=0 \\ 2\cos 2x−\cos x−1=&0 \\ (2\cos x−1)(\cos x+1)&=0 \end{aligned}

Set each factor equal to zero and solve.

\begin{aligned} 2\cos x−1&=0 \\ 2\cos x&=1 & \cos x+1 &=0\\ \cos x&=\dfrac{1}{2} &\text{ and} \qquad \cos x&=−1\\ x&=\dfrac{\pi }{3},\; \dfrac{5\pi }{3} & x&=\pi \end{aligned}

### Review

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

1. $$\cos x−\cos \dfrac{1}{2} x=0$$
2. $$\sin 2x\cos x=\sin x$$
3. $$\cos 3x−\cos ^3x=3\sin ^2x\cos x$$
4. $$\tan 2x−\tan x=0$$
5. $$\cos 2x−\cos x=0$$
6. $$2\cos ^2\dfrac{x}{2}=1$$
7. $$\tan \dfrac{x}{2}=4$$
8. $$\cos \dfrac{x}{2}=1+\cos x$$
9. $$\sin 2x+\sin x=0$$
10. $$\cos ^2x−\cos 2x=0$$
11. $$\dfrac{\cos 2x}{\cos ^2x}=1$$
12. $$\cos 2x−1=\sin ^2x$$
13. $$\cos 2x=\cos x$$
14. $$\sin 2x−\cos 2x=1$$
15. $$\sin ^2x−2=\cos 2x$$
16. $$\cot x+\tan x=2\csc 2x$$