Trigonometric Functions — Maths STD 12 Science — Question

$\therefore \quad \sin \theta=\sin \frac{4 \pi}{3}\left(\right.$ As $\sin \frac{4 \pi}{3}=\frac{\sqrt{3}}{2}$ and $\left.\sin (\pi+ A )=-\sin A \right)$

The general solution of $\sin \theta=\sin \alpha$ is $\alpha= n \pi+(-1)^{ n } \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sin \theta=\sin \frac{4 \pi}{3}$ is $\theta=n \pi+(-1)^{ n } \frac{4 \pi}{3}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sin \theta=-\frac{\sqrt{3}}{2}$ is $\theta=n \pi+(-1)^{ n } \frac{4 \pi}{3}$, where $n \in Z$.