Question

In Problems 79-90, find the exact value of each expression.

$f\left(x\right)=\cos \left(2x\right)+\cos x$

Short answer

The real zeroes are$\mathrm{\pi },\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}$

Step-by-step solution

Given information

Given function$f\left(x\right)=\cos \left(2x\right)+\cos x$

Rearranging and calculating the value

Rearranging, we get

$$\begin{gathered} f\left(x\right)=\cos \left(2x\right)+\cos x \\ f\left(x\right)=2{\cos }^{2}x-1+\cos x \\ (\cos x+1)(2\cos x-1)=0 \\ \cos x=-1 \\ \cos x=\frac{1}{2} \\ x=\pi ,\frac{\pi }{3},\frac{5\pi }{3} \end{gathered}$$