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

Question:

Find the general solution of each of the following equations:

(i) $\cos x=\frac{-1}{2}$

(ii) $\operatorname{cosec} x=-\sqrt{2}$

(iii) $\tan x=-1$

Solution:

To Find: General solution.

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

By using above formula, we have

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

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

(ii) Given: $\operatorname{cosec} x=-\sqrt{2}$

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

So $\sin x=\frac{-1}{\sqrt{2}}$

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

By using above formula, we have

$\sin x=\frac{-1}{\sqrt{2}}=\sin \frac{5 \pi}{4} \Rightarrow x=n \pi+(-1)^{n} \cdot \frac{5 \pi}{4}$

So general solution is $x=n \pi+(-1)^{n} \cdot \frac{5 \pi}{4}$ where $n \in I$

(iii) Given: $\tan x=-1$

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

By using above formula, we have

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

So the general solution is $x=n \pi+\frac{3 \pi}{4}$ where $n \in$ ।