Solve: \sec x-\tan x=\sqrt{3} .
We have
Thus, the given equation becomes
\sqrt{3} \cos x+\sin x=1.\qquad \qquad …(i)Dividing both sides of (i) by \sqrt{(\sqrt{3})^{2}+1^{2}} , i.e., by 2 , we get
Hence, the general solution is x=\left(2 n \pi-\frac{\pi}{6}\right) , where n \in I .