Solve the trigonometric equation: - \[\sin \theta +\sin 5\theta =\sin 3\theta \].

Hint: Use the sum to product rule of conversion of trigonometric function given by: - \[\sin a+\sin b=2\sin \dfrac{a+b}{2}\cos \dfrac{a-b}{2}\] to simplify the L.H.S. Take all the terms to the L.H.S to get 0 in the R.H.S. Take common terms together and write the equation as a product of two terms. Substitute each term equal to 0 and find the general solution. Use: - if \[\sin a=0\Rightarrow a=n\pi \] and if \[\cos a=\cos b,\Rightarrow a=2n\pi +b\], where \[n\in I\].

Complete step-by-step solution
We have been provided with the trigonometric equation: - \[\sin \theta +\sin 5\theta =\sin 3\theta \].
Using the identity: - \[\sin a+\sin b=2\sin \dfrac{a+b}{2}\cos \dfrac{a-b}{2}\] we get,
\[\begin{align}
  & \Rightarrow 2\sin \left( \dfrac{\theta +5\theta }{2} \right)\cos \left( \dfrac{\theta -5\theta }{2} \right)=\sin 3\theta \\
 & \Rightarrow 2\sin 3\theta \cos \left( -2\theta \right)=\sin 3\theta \\
\end{align}\]
We know that, \[\cos \left( -x \right)=\cos x\],
\[\begin{align}
  & \Rightarrow 2\sin 3\theta \cos 2\theta =\sin 3\theta \\
 & \Rightarrow 2\sin 3\theta \cos 2\theta -\sin 3\theta =0 \\
\end{align}\]
Taking \[\sin 3\theta \] common we get,
\[\Rightarrow \sin 3\theta \left( 2\cos 2\theta -1 \right)=0\]
Substituting each term equal to 0 one – by – one we get,
Case (i): - When \[\sin 3\theta =0\],
We know that, if \[\sin a=0\Rightarrow a=n\pi \], \[n\in I\]
\[\begin{align}
  & \Rightarrow \sin 3\theta =0 \\
 & \Rightarrow 3\theta =n\pi \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{n\pi }{3},n\in I\]
Case (ii): - When \[\left( 2\cos 2\theta -1 \right)=0\],
\[\begin{align}
  & \Rightarrow 2\cos 2\theta =1 \\
 & \Rightarrow \cos 2\theta =\dfrac{1}{2} \\
 & \Rightarrow \cos 2\theta =\cos \dfrac{\pi }{3} \\
\end{align}\]
We know that, if \[\cos a=\cos b,\Rightarrow a=2n\pi +b\], \[n\in I\].
\[\begin{align}
  & \Rightarrow \cos 2\theta =\cos \dfrac{\pi }{3} \\
 & \Rightarrow 2\theta =2n\pi +\dfrac{\pi }{3} \\
 & \Rightarrow \theta =n\pi +\dfrac{\pi }{6} \\
 & \Rightarrow \theta =\left( 6n+1 \right)\dfrac{\pi }{6},n\in I \\
\end{align}\]

Note: One may note that we have to consider both the cases as our solution because both will satisfy the equation. Remember that we do not have to cancel \[\sin 3\theta \] from both sides, if we will do so then there will be loss of many roots. So, instead of canceling, we have to take common and consider each condition. Always remember the basic formula to write general solutions of all the trigonometric functions.