Answer
(a) $ x=\frac{\pi}{3}+2n\pi,\ \ x= \frac{5\pi}{3}+2n\pi ,\ \text{or}\ \pi +2n\pi $
(b) $ \bigg\{ \frac{\pi}{3} , \ \pi ,\frac{3\pi}{3} \bigg\} $
Work Step by Step
(a) Given
$$2\cos^2 x+\cos x-1 =0 $$
Since
\begin{align*}
2\cos^2 x+\cos x-1 &=0\\
(2\cos x-1)(\cos x+1)& =0
\end{align*}
Then
$$2\cos x-1=0,\ \ \text{or} \ \ \cos x+1=0$$
Case 1 , $$\cos x=\frac{1}{2},\ \ \Rightarrow \ \ x=\frac{\pi}{3}+2n\pi,\ \ \text{or}\ \ x= \frac{5\pi}{3}+2n\pi $$
Case 2 , $$\cos x =-1,\ \ \Rightarrow \ \ x=\pi +2n\pi $$
Hence the general solutions is
$$ x=\frac{\pi}{3}+2n\pi,\ \ x= \frac{5\pi}{3}+2n\pi ,\ \text{or}\ \pi +2n\pi $$
(b) To find solutions on the interval $ 0\leq x<2\pi $, put values of $n $
$$ \bigg\{ \frac{\pi}{3} , \ \pi ,\frac{3\pi}{3} \bigg\} $$
