## Calculus 8th Edition

$\frac{ \pi}{2},\frac{3 \pi}{2}$
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, $\frac{ \pi}{2},\frac{3 \pi}{2}$