Answer
(a) $ x= \frac{\pi}{2}+n\pi ,\ \ \ x=\frac{\pi}{6} +2n\pi,\ \ \text{or} \frac{5\pi}{6} +2n\pi$
(b) $ \bigg\{\frac{\pi}{6} , \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{6} \bigg\} $
Work Step by Step
(a) Given
$$\cos x-2\sin x\cos x =0 $$
Since
\begin{align*}
\cos x-2\sin x\cos x &=0\\
\cos x(1- 2\sin x)& =0
\end{align*}
Then
$$\cos x=0,\ \ \text{or} \ \ 1- 2\sin x=0$$
Case 1 , $$\cos x=0,\ \ \Rightarrow \ \ x=\frac{\pi}{2}+n\pi $$
Case 2 , $$\sin x =\frac{1}{2},\ \ \Rightarrow \ \ x=\frac{\pi}{6} +2n\pi,\ \ \text{or} \frac{5\pi}{6} +2n\pi$$
Hence the general solutions is
$$ x= \frac{\pi}{2}+n\pi ,\ \ \ x=\frac{\pi}{6} +2n\pi,\ \ \text{or} \frac{5\pi}{6} +2n\pi$$
(b) To find solutions on the interval $ 0\leq x<2\pi $, put values of $n $
$$ \bigg\{\frac{\pi}{6} , \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{6} \bigg\} $$
