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

Question:

Find the general solution of each of the following equations:

$\sin x \tan x-1=\tan x-\sin x$

Solution:

To Find: General solution.

Given: $\sin x \tan x-1=\tan x-\sin x \Rightarrow \sin x(\tan x+1)=\tan x+1$

So $\sin x=1=\sin \left(\frac{\pi}{2}\right)$ or $\tan x=-1=\tan \left(\frac{3 \pi}{4}\right)$

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

$\Rightarrow x=n \pi+(-1)^{n} \frac{\pi}{2}$ or $x=k \pi+\frac{3 \pi}{4}$ where $n, k \in I$

So general solution is $x=n \pi+(-1)^{n} \frac{\pi}{2}$ or $x=k \pi+\frac{3 \pi}{4}$ where $n, k \in$ ।