Question

Find the general solution of \[\theta \]. If \[\sin 3\theta = \sin \theta \]
A) \[2n\pi ,(2n + 1)\dfrac{\pi }{3}\]
B) \[n\pi ,(2n + 1)\dfrac{\pi }{4}\]
C) \[n\pi ,(2n + 1)\dfrac{\pi }{3}\]
D) None of these

Answer 1

Hint: If \[\sin \alpha = \sin \beta \] then it must be known that one of the angles is just completed a whole \[2n\pi \] rotation to reach there only i.e., \[\sin \dfrac{\pi }{3} = \sin \dfrac{{7\pi }}{3}\] because \[\sin \dfrac{{7\pi }}{3} = \sin \left( {2\pi + \dfrac{\pi }{3}} \right)\] .
Complete step by step Solution:
We know that when \[\sin \alpha = \sin \beta \]
\[\therefore \alpha = \beta + 2n\pi \] or \[\therefore \beta = \alpha + 2n\pi \]
Using this trick we can write
When \[\sin 3\theta = \sin \theta \]
\[\begin{array}{l}
\therefore 3\theta = \theta + 2n\pi \\
 \Rightarrow 2\theta = 2n\pi \\
 \Rightarrow \theta = n\pi
\end{array}\]
Now as we have one value for the other value we also know that
when \[\sin \alpha = \sin \beta \] then \[\alpha = \pi - \beta + 2n\pi \]
So from here we can write it as
\[\begin{array}{l}
\therefore \theta = \pi - 3\theta + 2n\pi \\
 \Rightarrow 4\theta = \pi + 2n\pi \\
 \Rightarrow \theta = \dfrac{\pi }{4}(2n + 1)
\end{array}\]
So now we have both the values
Therefore option B, \[n\pi ,(2n + 1)\dfrac{\pi }{4}\] is the correct option.

Note: We can also do this question by writing the formula of \[\sin 3\theta \] as \[3\sin \theta - 4{\sin ^3}\theta \] and then solve the equation in sine to get the final values.