Question

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

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

Short answer

The solution set for the equation$\cos \left(2\theta \right)+\cos \left(4\theta \right)=0$is,$\left\{\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2},\frac{11\mathrm{\pi }}{6}\right\}$.

Step-by-step solution

Step 1 Given equation is,

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

The double angle formula 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.

$$\begin{gathered} \cos \left(2\theta \right)+\left(2{\cos }^2\left(2\theta \right)-1\right)=0 \\ (2{\cos }^2\theta -1)+\left(2{\cos }^2\left(2\theta \right)-1\right)=0 \\ (2{\cos }^2\theta -1)+\left(2{(2{\cos }^2\theta -1)}^2-1\right)=0 \\ (2{\cos }^2\theta -1)+(2\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)-1)=0 \\ (2{\cos }^2\theta -1)+\left(8{\cos }^4\theta -8{\cos }^2\theta +1\right)=0 \\ 8{\cos }^4\theta -6{\cos }^2\theta =0 \\ 2{\cos }^2\theta \left(4{\cos }^2\theta -3\right)=0 \end{gathered}$$

Step 2 Apply the zero product property.

$$\begin{gathered} 2{\cos }^2\theta =0or\left(4{\cos }^2\theta -3\right)=0 \\ {\cos }^2\theta =0or4{\cos }^2\theta =3 \\ \cos \theta =0or{\cos }^2\theta =\frac{3}{4} \\ \cos \theta =\pm \sqrt{\frac{3}{4}} \\ \cos \theta =\pm \frac{\sqrt{3}}{2} \end{gathered}$$

Solving each equation in the interval $[0,2\mathrm{\pi })$is,

$$\begin{gathered} \theta =\frac{\mathrm{\pi }}{2} \\ \theta =\frac{3\mathrm{\pi }}{2} \\ \theta =\frac{\mathrm{\pi }}{6} \\ \theta =\frac{11\mathrm{\pi }}{6} \\ \theta =\frac{5\mathrm{\pi }}{6} \\ \theta =\frac{7\mathrm{\pi }}{6} \end{gathered}$$

Therefore, the solution set is,$\left\{\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2},\frac{11\mathrm{\pi }}{6}\right\}$.