Question

In Problems $69-78$, solve each equation on the interval $0\le \theta <2\mathrm{\pi }$.

$\sin \left(2\theta \right)+\sin \left(4\theta \right)=0$.

Short answer

The solution set for the equation$\sin \left(2\theta \right)+\sin \left(4\theta \right)=0$is,$\left\{0,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2},\mathrm{\pi },\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{3}\right\}$.

Step-by-step solution

Step 1Given equation is,

$\sin \left(2\theta \right)+\sin \left(4\theta \right)=0$.

The double angle for $\sin \left(2\theta \right)$is, $\sin \left(2\theta \right)=2\sin \theta \cos \theta$and for $\cos \left(2\theta \right)$is,$\cos \left(2\theta \right)=2{\cos }^2\theta -1$.

Substitute the known values in the given equation.

$$\begin{gathered} \sin \left(2\theta \right)+2\sin \left(2\theta \right)\cos \left(2\theta \right)=0 \\ \sin \left(2\theta \right)\left(1+2\cos \left(2\theta \right)\right)=0 \\ \left[2\sin \theta \cos \theta \right]\left(1+2\left(2{\cos }^2\theta -1\right)\right)=0 \\ \left[2\sin \theta \cos \theta \right]\left(1+4{\cos }^2\theta -2\right)=0 \\ \left[2\sin \theta \cos \theta \right]\left(4{\cos }^2\theta -1\right)=0 \end{gathered}$$

Step 2 Apply the zero product property.

$$\begin{gathered} \left[2\sin \theta \cos \theta \right]=0or\left(4{\cos }^2\theta -1\right)=0 \\ \sin \theta \cos \theta =0or4{\cos }^2\theta =1 \\ \sin \theta \cos \theta =0or{\cos }^2\theta =\frac{1}{4} \\ \sin \theta \cos \theta =0or\cos \theta =\pm \frac{1}{2} \\ \sin \theta =0or\cos \theta =0or\cos \theta =\pm \frac{1}{2} \end{gathered}$$

Step 3 Solving the three equations in the interval [0,2π) we get 7 values for θ.

$\theta =0,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2},\frac{2\mathrm{\pi }}{3},\mathrm{\pi },\frac{4\mathrm{\pi }}{3},\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{3}$.

Therefore the solution set is,$\left\{0,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2},\frac{2\mathrm{\pi }}{3},\mathrm{\pi },\frac{4\mathrm{\pi }}{3},\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{3}\right\}$.