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 $\sin 3\theta = \sin \theta$

\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.