## Calculus 8th Edition

$0,\frac{\pi}{3},\frac{5 \pi}{3}, 2 \pi$
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, $0,\frac{\pi}{3},\frac{5 \pi}{3}, 2 \pi$