Question

Find the number of distinct solution of $\sec x+\tan x=\sqrt{3},$ where $0\le x\le 3\pi$

Answer 1

Here,$\sec x+\tan x=\sqrt{3}$

$\Rightarrow \frac{1}{\cos x}+\frac{\sin x}{\cos x}=\sqrt{3}$

$\Rightarrow 1+\sin x=\sqrt{3}\cos x$

$\Rightarrow \sqrt{3}\cos x-\sin x=1$ is of the form $a\cos x+b\sin x=c$

where $a=\sqrt{3},b=-1$

Dividing both sides by $\sqrt{a^2+b^2}=\sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2}=\sqrt{4}=2$, we get

$\Rightarrow \frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=\frac{1}{2}$

$\Rightarrow \cos \frac{\pi }{6}\cos x-\sin \frac{\pi }{6}\sin x=\cos \frac{\pi }{3}$

$\Rightarrow \cos (x+\frac{\pi }{6})=\cos \frac{\pi }{3}$

$\Rightarrow x+\frac{\pi }{6}=2n\pi \pm \frac{\pi }{3}$

$\Rightarrow x=2n\pi \pm \frac{\pi }{3}-\frac{\pi }{6}$

$\Rightarrow x=2n\pi +\frac{\pi }{6}$ or $x=2n\pi -\frac{\pi }{2}$

there are no solution for $n=0,1$ and also for $x=2n\pi -\frac{\pi }{2}$

$\therefore$ Total number of solutions are $2$