AMC 12 Mock Exam 2 Problem 14

By the sum-to-product formula, $\cos x+\cos5x=2\cos3x\cos2x$, so the equation becomes $\cos3x\,(2\cos2x+1)=0$.

$\cos3x=0$: with $3x\in[0,3\pi]$, $3x\in\left\{\frac\pi2,\frac{3\pi}2,\frac{5\pi}2\right\}$, so $x\in\left\{\frac\pi6,\frac\pi2,\frac{5\pi}6\right\}$.

$\cos2x=-\frac12$: with $2x\in[0,2\pi]$, $2x\in\left\{\frac{2\pi}3,\frac{4\pi}3\right\}$, so $x\in\left\{\frac\pi3,\frac{2\pi}3\right\}$.

These five values are distinct, so there are $\boxed{5}$ solutions.