Answer
$x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=\pi + 2 \pi n$
Work Step by Step
Here, we have $ 2 \cos^2 x-1+\cos x=0$
This gives: $ (2 \cos x-1) (1+\cos x)=0$
The general solution for $\cos (a)=\cos (0)$ is $ a=2 n \pi $
$ 2 \cos x-1=0 \implies \cos x=\dfrac{1}{2}$
This gives: $ x=\dfrac{\pi}{3}; \dfrac{ 5\pi}{3}$
and $ 1+ \cos x=0 \implies \cos x=-1$
This gives: $ x=\pi $
Hence, we have $x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=\pi + 2 \pi n$
