find all solutions to cos(9t) - cos(3t) = sin(6t) on 0 ≤ t < 2π/3

Explanation:

Step1: Use the sum - to - product formula

We know that $\cos A-\cos B=- 2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)$. Here $A = 9t$ and $B=3t$, so $\cos(9t)-\cos(3t)=-2\sin(6t)\sin(3t)$. The equation becomes $-2\sin(6t)\sin(3t)=\sin(6t)$.

Step2: Rearrange the equation

Move all terms to one side: $\sin(6t)+2\sin(6t)\sin(3t)=0$. Factor out $\sin(6t)$: $\sin(6t)(1 + 2\sin(3t))=0$.

Step3: Solve for $\sin(6t)=0$

If $\sin(6t)=0$, then $6t = k\pi$, where $k\in\mathbb{Z}$. So $t=\frac{k\pi}{6}$. Since $0\leq t<\frac{2\pi}{3}$, when $k = 0$, $t = 0$; when $k=1$, $t=\frac{\pi}{6}$; when $k = 2$, $t=\frac{\pi}{3}$; when $k=3$, $t=\frac{\pi}{2}$; when $k = 4$, $t=\frac{2\pi}{3}$ (but this is not in the domain, so we discard it).

Step4: Solve for $1 + 2\sin(3t)=0$

If $1+2\sin(3t)=0$, then $\sin(3t)=-\frac{1}{2}$. So $3t=\frac{7\pi}{6}+2k\pi$ or $3t=\frac{11\pi}{6}+2k\pi$, $k\in\mathbb{Z}$. For $3t=\frac{7\pi}{6}+2k\pi$, $t=\frac{7\pi}{18}+\frac{2k\pi}{3}$; for $3t=\frac{11\pi}{6}+2k\pi$, $t=\frac{11\pi}{18}+\frac{2k\pi}{3}$. In the domain $0\leq t<\frac{2\pi}{3}$, when $k = 0$, $t=\frac{7\pi}{18}$ and $t=\frac{11\pi}{18}$.

Answer:

$t = 0,\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{7\pi}{18},\frac{11\pi}{18}$