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