$\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . Find number of solutions and the solutions for this equation in $[0,\pi]$ I tried to solve the equation by doing this-
$$2\sin(\frac{\theta+5\theta}{2})\cos(\frac{\theta-5\theta}{2})=\sin(3\theta)\\
  2\sin(3\theta)\cos(2\theta) = \sin(3θ)\\
  \cos(2θ) = \frac{1}{2}\\
  1-2\sin^2(θ) = \frac{1}{2}\\
\sin^2(θ) = (\frac{1}{2})^2\\
∴ θ = nπ ± α\\
Answer = \frac{π}{6},\frac{5π}{6} $$
But in the solution there are 6 solutions and in step 3 instead of dividing $\sin(3θ)$ by $\sin(3θ)$ they have taken it as common and made "$\sin(3θ)(2\cos(2θ)-1)$"

As we cannot divide by $0$, there are two possible cases:
(1) $\sin3\theta=0$ (i.e. $\theta=0$, $\frac \pi3$, $\frac {2\pi}3$ or $\pi$)
Then both $2\sin3\theta\cos2\theta$ and $\sin3\theta$ are $0$. The equation is satisfied. We have $4$ solutions in this case.
(2) $\sin3\theta\ne0$.
Then we can divide both sides of $2\sin3\theta\cos2\theta=\sin3\theta$ by $\sin3\theta$ as it is not zero. So we have $\cos2\theta=\frac12$ and hence $\theta=\frac \pi6$ or $\frac{5\pi}6$. There are $2$ solutions in this case.
