3.4.4: Solving Equations with Double-Angle Identities

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.

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

2. 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 \(x=\dfrac{4 \pi}{3}\).

3. 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. \(\cot x+\tan x=2\csc 2x\)

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.17.