AMC 12 Mock Exam 3 Problem 24

Repeated use of $2\sin\theta\cos\theta=\sin2\theta$ gives $16\sin x\cos x\cos2x\cos4x\cos8x=\sin16x$.

Points with $\sin x=0$: at $x=0,2\pi$ the product is $1$, and at $x=\pi$ it is $-1$; none is a solution.

Otherwise the equation is equivalent to $\sin16x=\sin x$, i.e., either $16x=x+2\pi k$ or $16x=\pi-x+2\pi k$:

• $x=\frac{2\pi k}{15}$: in $(0,2\pi)$ these are $k=1,\dots,14$ (none equals $\pi$): $14$ solutions.
• $x=\frac{(2k+1)\pi}{17}$: in $(0,2\pi)$ these are $k=0,\dots,16$, excluding $k=8$ (which gives $x=\pi$): $16$ solutions.

The families are disjoint, because $\frac{2k}{15}=\frac{2j+1}{17}$ would give $34k=15(2j+1)$, even $=$ odd. The total is $14+16=\boxed{30}$.