AMC 12 Mock Exam 1 Problem 13

Since $\cos3x=\sin\left(\frac\pi2-3x\right)$, the equation is $\sin 2x=\sin\left(\frac\pi2-3x\right)$, which holds exactly when either

(i) $2x=\frac\pi2-3x+2\pi k$, i.e., $x=\frac{\pi}{10}+\frac{2\pi k}{5}$, or

(ii) $2x=\pi-\left(\frac\pi2-3x\right)+2\pi k$, i.e., $x=-\frac\pi2-2\pi k$.

In $[0,2\pi)$, family (i) gives the five values $\frac{\pi}{10},\frac{\pi}{2},\frac{9\pi}{10},\frac{13\pi}{10},\frac{17\pi}{10}$ (for $k=0,\dots,4$), whose sum is $\frac{\pi}{10}(1+5+9+13+17)=\frac{9\pi}{2}$. Family (ii) gives the single value $x=\frac{3\pi}{2}$. These six values are distinct, so the sum is $\frac{9\pi}2+\frac{3\pi}2=\boxed{6\pi}$.