Question

If $\sin\theta = \sin\left( \frac{- \pi}{6} \right)$, then the general value of $\Rightarrow$ is.

• A. $\theta = n\pi + ( - 1)^{n}\left( \frac{- \pi}{6} \right) \Rightarrow \theta = n\pi + ( - 1)^{n + 1}\frac{\pi}{6}$
• B. $\Rightarrow$
• C. $\theta = n\pi + ( - 1)^{n}\frac{7\pi}{6}$
• D. None of these

Answer 1

Answer: (C)

$\theta = n\pi + ( - 1)^{n}\frac{7\pi}{6}$

Answer 2

$\Rightarrow$

$\sin 4\theta = 0 \Rightarrow 4\theta = n\pi$

Hence$\sin 3\theta = \frac{1}{2} = \sin\left( \frac{\pi}{6} \right)$ or $\Rightarrow$.

But $3\theta = n\pi + ( - 1)^{n}\frac{\pi}{6} \Rightarrow \theta = \frac{n\pi}{4},\frac{n\pi}{3} + ( - 1)^{n}\frac{\pi}{18}$ is ruled out.