Answer
$\frac{ \pi}{2},\frac{3 \pi}{2}$,$\frac{\pi}{6},\frac{5 \pi}{6}$
Work Step by Step
Need to find the range for $x$ for the equation
$sin2x=cosx$
$2sinxcosx-cosx=0$
$cosx(2sinx-1)=0$
Here, $cosx=0$ gives $x =\frac{ \pi}{2},\frac{3 \pi}{2}$
and $2sinx-1=0$ gives $sinx=\frac{1}{2}$
Since, $sin>0$ on first and second quadrant , thus, x must lie on $\frac{\pi}{6},\frac{5 \pi}{6}$
Hence, $\frac{ \pi}{2},\frac{3 \pi}{2}$,$\frac{\pi}{6},\frac{5 \pi}{6}$
