$\sin(\theta) + \sin(5\theta) = \sin(3\theta)$ . 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 "$\sin(3θ)(2\cos(2θ)-1)$"
 A: That's a very common mistake: if you have $ab=a$, you cannot conclude that $b=1$, unless you know that $a\ne0$. Indeed, you can rewrite the relation as
$$
a(b-1)=0
$$
and therefore either $a=0$ or $b=1$.
So from $2\sin3\theta\cos2\theta=\sin3\theta$ you cannot draw just $2\cos2\theta=1$: the procedure of your textbook is correct:
$$
2\sin3\theta(\cos2\theta-1)=0
$$
so
$$
\sin3\theta=0 \qquad\text{or}\qquad 2\cos2\theta-1=0
$$
Now the solutions.
From $\sin3\theta=0$ we get $3\theta=n\pi$, so $\theta=n\pi/3$ and there are four solutions in $[0,\pi]$, namely $0$, $\pi/3$, $2\pi/3$ and $\pi$.
From $\cos2\theta=1/2$ we get either $2\theta=\pi/3+2n\pi$ or $2\theta=-\pi/3+2n\pi$, hence
$$
\theta=\frac{\pi}{6}+n\pi
\qquad\text{or}\qquad
\theta=-\frac{\pi}{6}+n\pi
$$
getting also the solutions $\pi/6$ and $5\pi/6$.
A: 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.
A: $$\sin3\theta=2\sin\dfrac{5\theta+\theta}2\cos\dfrac{5\theta-\theta}2$$
$$\implies\sin3\theta(2\cos2\theta-1)=0$$
If $\sin3\theta=0,3\theta=m\pi$ where $m$ is any integer
We need $0\le\dfrac{m\pi}3\le\pi\iff 0\le m\le3$
If $2\cos2\theta-1=0,\sin^2\theta=\sin^2\dfrac{\pi}6$
$\theta=n\pi\pm\dfrac{\pi}6$
We need $0\le n\pi\pm\dfrac{\pi}6\le\pi\iff0\le 6n\pm1\le6$
Taking '+' sign, $0\le 6n+1\le6\implies-1<-\dfrac16\le n\le\dfrac56<1\implies n=0$
Take '-' sign
