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# Solving Equations with Double-Angle Identities

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

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Practice Solving Equations with Double-Angle Identities
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Solving Trig Equations using Double and Half Angle Formulas

Trig Riddle #4: I am an angle x such that $0\le x <2\pi$ . I satisfy the equation $\sin 2x - \sin x=0$ . What angle am I?

### Guidance

Lastly, we can use the half and double angle formulas to solve trigonometric equations.

#### Example A

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

Solution: Change $\tan 2x$ and simplify.

$\tan 2x + \tan x &=0 \\\frac{2\tan x}{1-\tan ^2 x}+\tan x &=0 \\2\tan x +\tan x(1-\tan ^2x) &=0 \quad \rightarrow \text{Multiply everything by} \ 1-\tan^2x \text{ to eliminate denominator.}\\2\tan x +\tan x -\tan ^3 x &=0 \\3\tan x - \tan ^3 x &=0 \\\tan x(3-\tan^2 x) &=0$

Set each factor equal to zero and solve.

$&\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3-\tan ^2x =0 \\&\ \ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ -\tan ^2 x =-3 \\&\tan x=0 \qquad \qquad \qquad and \qquad \qquad \ \tan^2 x=3 \\& \qquad x=0 \ and \ \pi \ \qquad \qquad \qquad \quad \qquad \tan x =\pm \sqrt{3} \\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x =\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$

#### Example B

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

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

$2\cos \frac{x}{2}+1 &=0 \\2\cos \frac{x}{2} &=-1 \\\cos \frac{x}{2} &= -\frac{1}{2}$

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

$\frac{x}{2} &=\frac{2\pi}{3},\frac{4\pi}{3} \\&=\frac{4\pi}{3}, \frac{8\pi}{3}$

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

#### Example C

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

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

$4\sin x \cos x &=\sqrt{3} \\2\cdot 2\sin x \cos x &=\sqrt{3} \\2 \cdot \sin 2x &=\sqrt{3} \\\sin 2x &=\frac{\sqrt{3}}{2} \\2x &= \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3} \\x &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

Concept Problem Revisit

Use the double angle formula and simplify.

$\sin 2x - \sin x = 0\\2\sin x \cos x - \sin x = 0\\\sin x(2\cos x - 1)= 0\\\sin x = 0 \text OR \cos x = \frac{1}{2}$

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

### Guided Practice

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

1. $\sin \frac{x}{2}=-1$

2. $\cos 2x-\cos x=0$

1. $\sin \frac{x}{2} &=-1 \\\frac{x}{2} &=\frac{3\pi}{2} \\x &=3\pi$

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

2. $\cos 2x - \cos x &=0 \\2\cos ^2x-\cos x -1 &=0 \\(2\cos x -1)(\cos x +1) &=0$

Set each factor equal to zero and solve.

$2\cos x-1 &=0 \\2\cos x &=1 \qquad \qquad \qquad \qquad\cos x +1=0\\\cos x &=\frac{1}{2} \qquad \qquad and \qquad \qquad \cos x =-1\\x &=\frac{\pi}{3}, \frac{5\pi}{3} \qquad \qquad \qquad \qquad \quad \ x=\pi \\$

### Practice

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

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