Find the general solution of each of the following equations:

Solution:

To Find: General solution.

(i) Given: $\sin 2 x=\frac{1}{2}$

Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^{n} \alpha, n \in I$

By using above formula, we have

$\sin 2 x=\frac{1}{2}=\sin \frac{\pi}{6} \Longrightarrow 2 x=n \pi+(-1)^{n} \cdot \frac{\pi}{6} \Longrightarrow x=\frac{n \pi}{2}+(-1)^{n} \cdot \frac{\pi}{12}, n \in I$

So general solution is $x=\frac{n \pi}{2}+(-1)^{n} \cdot \frac{\pi}{12}$ where $n \in I$

(ii) Given: $\cos 3 x=\frac{1}{\sqrt{2}}$

Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in I$

By using above formula, we have

$\cos 3 x=\frac{1}{\sqrt{2}}=\cos \left(\frac{\pi}{4}\right) \Rightarrow 3 x=2 n \pi \pm \frac{\pi}{4} \Rightarrow x=\frac{2 n \pi}{3} \pm \frac{\pi}{12}, n \in I$

So the general solution is $x=\frac{2 n \pi}{3} \pm \frac{\pi}{12}$ where $n \in$ ।

(iii) Given: $\tan \frac{2 x}{3}=\sqrt{3}$

Formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=\mathrm{n} \pi+\alpha, \mathrm{n} \in$ ।

By using above formula, we have

$\tan \frac{2 \mathrm{x}}{3}=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \frac{2 \mathrm{x}}{3}=\mathrm{n} \pi+\frac{\pi}{3} \Rightarrow \mathrm{x}=\frac{3 \mathrm{n} \pi}{2}+\frac{\pi}{2}, \mathrm{n} \in I$

So general solution is $x=(3 n+1) \frac{\pi}{2}$, where $n \in I$