# Find the general solution of each of the following equations:

Question:

Find the general solution of each of the following equations:

(i) $\sec 3 x=-2$

(ii) $\cot 4 x=-1$

(iii) $\operatorname{cosec} 3 x=\frac{-2}{\sqrt{3}}$

Solution:

To Find: General solution.

(i) Given: $\sec 3 x=-2$

We know that $\sec \theta \times \cos \theta=1$

So $\cos 3 x=\frac{-1}{2}$

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

By using above formula, we have

$\cos 3 x=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} \Rightarrow 3 x=2 n \pi \pm \frac{2 \pi}{3} \Rightarrow x=\frac{2 n \pi}{3} \pm \frac{2 \pi}{9}$ $n \in I$

So the general solution is $x=\frac{2 n \pi}{3} \pm \frac{2 \pi}{9}$, where $n \in I$

(ii) Given: $\cot 4 x=-1$

We know that $\tan \theta \times \cot \theta=1$

So $\tan 4 x=-1$

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

By using above formula, we have

$\tan 4 x=-1=\tan \frac{3 \pi}{4} \Rightarrow 4 x=n \pi+\frac{3 \pi}{4} \Rightarrow x=\frac{n \pi}{4}+\frac{3 \pi}{16}, n \in I$

So general solution is $x=(4 n+3) \frac{\pi}{16}$, where $n \in I$

(iii) Given: $\operatorname{cosec} 3 x=\frac{-2}{\sqrt{3}}$

We know that $\operatorname{cosec} \theta \times \sin \theta=1$

So $\sin 3 x=\frac{-\sqrt{3}}{2}$

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

By using above formula, we have

$\sin 3 x=\frac{-\sqrt{3}}{2}=\sin \frac{4 \pi}{3} \Rightarrow 3 x=n \pi+(-1)^{n} \cdot \frac{4 \pi}{3} \Rightarrow x=\frac{n \pi}{3}+(-1)^{n} \cdot \frac{4 \pi}{9}, n \in I$

So general solution is $x=\frac{n \pi}{3}+(-1)^{n} \cdot \frac{4 \pi}{9}$, where $n \in I$