Answer
$0,\frac{\pi}{3},\frac{5 \pi}{3}, 2 \pi$
Work Step by Step
Need to find the range for $x$ for the equation
$2+cos2x=3cosx$
$2+(2cos^{2}x-1)=3cosx$
$2cos^{2}x-3cosx+1=0$
$(2cosx-1)$ $(cosx-1)=0$
Here, $2cosx-1=0$ gives
$cosx=\frac{1}{2}$
Since, $cosx>0$ on first and fourth quadrant thus, x must lie on $\frac{\pi}{3},\frac{5 \pi}{3}$
and $(cosx-1)=0$ gives $cosx=1$
$x =0, 2\pi$
Hence, $0,\frac{\pi}{3},\frac{5 \pi}{3}, 2 \pi$
