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 "$\sin(3θ)(2\cos(2θ)-1)$"
Best Answer
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 $5\pi/6$.