Find the general solution of each of the following equations:

Solution:

To Find: General solution.

[NOTE: A solution of a trigonometry equation generalized by means of periodicity, is known as general solution]

(i) Given: $\sin 3 x=0$

Formula used: $\sin \theta=0 \Rightarrow \theta=\mathrm{n} \pi, \mathrm{n} \in_{I}$

By using above formula, we have

$\sin 3 x=0 \Rightarrow 3 x=n \pi \Rightarrow x=\frac{n \pi}{3}$ where $n \in I$

So general solution is $x=\frac{n \pi}{3}$ where $n \in I$

(ii) Given: $\sin \frac{3 x}{2}=0$

Formula used: $\sin \theta=0 \Longrightarrow \theta=n \pi, n \in I$

By using above formula, we have

$\sin \frac{3 x}{2}=0 \Rightarrow \frac{3 x}{2}=\mathrm{n} \pi \Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3}$ where $\mathrm{n} \in$

So general solution is $x=\frac{2 n \pi}{3}$ where $n \in \mid$

(iii) Given: $\sin \left(x+\frac{\pi}{5}\right)=0$

Formula used: $\sin \theta=0 \Longrightarrow \theta=n \pi, n \in$ ।

By using the above formula, we have

$\sin \left(x+\frac{\pi}{5}\right)=0 \Rightarrow x+\frac{\pi}{5}=\mathrm{n} \pi \Rightarrow \mathrm{x}=\mathrm{n} \pi-\frac{\pi}{5}$ where $\mathrm{n} \in \mathrm{I}$

So general solution is $x=n \pi-\frac{\pi}{5}$ where $n \in I$

(iv) Given: $\cos 2 x=0$

Formula used: $\cos \theta=0 \Rightarrow \theta=(2 n+1) \frac{\pi}{2}, n \in I$

By using above formula, we have

$\cos 2 x=0 \Rightarrow 2 x=(2 n+1) \frac{\pi}{2} \Rightarrow x=(2 n+1) \frac{\pi}{4}$ where $n \in I$

So general solution is $x=(2 n+1) \frac{\pi}{4}$ where $n \in I$

(v) Given: $\cos \frac{5 x}{2}=0$

Formula used: $\cos \theta=0 \Longrightarrow \theta=(2 n+1) \frac{\pi}{2}, n \in I$

By using the above formula, we have

$\cos \frac{5 x}{2}=0 \Longrightarrow \frac{5 x}{2}=(2 n+1) \frac{\pi}{2} \Rightarrow x=(2 n+1) \frac{\pi}{5}$ where $n \in I$

So general solution is $x=(2 n+1) \frac{\pi}{5}$ where $n \in I$

(vi) Given: $\cos \left(x+\frac{\pi}{10}\right)=0$

Formula used: $\cos \theta=0 \Rightarrow \theta=(2 n+1) \frac{\pi}{2}, n \in I$

By using the above formula, we have

$\cos \left(x+\frac{\pi}{10}\right)=0 \Rightarrow x+\frac{\pi}{10}=(2 n+1) \frac{\pi}{2} \Rightarrow x=(2 n+1) \frac{\pi}{2}-\frac{\pi}{10} \Rightarrow x=n \pi+\frac{2 \pi}{5}$

where $n \in I$

So general solution is $\mathrm{x}=\mathrm{n} \pi+\frac{2 \pi}{5}$ where $\mathrm{n} \in$ ।

(vii) Given: $\tan 2 x=0$

Formula used: $\tan \theta=0 \Rightarrow \theta=n \pi, n \in I$

By using above formula, we have

$\tan 2 x=0 \Rightarrow 2 x=n \pi \Rightarrow x=\frac{n \pi}{2}$ where $n \in I$

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

(viii) Given: $\tan \left(3 x+\frac{\pi}{6}\right)=0$

Formula used: $\tan \theta=0 \Longrightarrow \theta=\mathrm{n} \pi, \mathrm{n} \in$ ।

By using above formula, we have

$\tan \left(3 x+\frac{\pi}{6}\right)=0 \Rightarrow 3 x+\frac{\pi}{6}=n \pi \Rightarrow 3 x=n \pi-\frac{\pi}{6} \Rightarrow x=\frac{n \pi}{3}-\frac{\pi}{18}$ where $n \in 1$

So general solution is $x=\frac{n \pi}{3}-\frac{\pi}{18}$ where $n \in I$

(ix) Given: $\tan \left(2 x-\frac{\pi}{4}\right)=0$

Formula used: $\tan \theta=0 \Longrightarrow \theta=n \pi, n \in I$

By using above formula, we have

$\tan \left(2 x-\frac{\pi}{4}\right)=0 \Rightarrow 2 x-\frac{\pi}{4}=n \pi \Rightarrow 2 x=n \pi-\frac{\pi}{4} \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{8}$ where $n \in I$

So general solution is $x=\frac{n \pi}{2}+\frac{\pi}{8}$ where $n \in$ ।