Question

The number of distinct solutions of $\sec x+\tan x=\sqrt{3}$, where $0\le x\le 3\pi$ is

Answer 1

Here, $\sec x+\tan x=\sqrt{3}$
$\Rightarrow 1+\sin x=\sqrt{3}\cos x$

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

dividing both sides by $\sqrt{a^2+b^2}$ i.e $\sqrt{4}=2$ , we get
$\Rightarrow \frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=\frac{1}{2}$
$\Rightarrow \cos \frac{\pi }{6}\cos x-\sin \frac{\pi }{6}\sin x=\frac{1}{2}$
$\Rightarrow \cos (x+\frac{\pi }{6})=\frac{1}{2}$
$\Rightarrow \cos (x+\frac{\pi }{6})=\cos \frac{\pi }{3}$
As $0\le x\le 3\pi$
$\frac{\pi }{6}\le x+\frac{\pi }{6}\le 3\pi +\frac{\pi }{6}$
$\Rightarrow x+\frac{\pi }{6}=\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3}$
$\Rightarrow x=\frac{\pi }{6},\frac{3\pi }{2},\frac{13\pi }{6}$
But at $x=\frac{3\pi }{2},\tan x\text{and}\sec x$ is not defined.

Alternate Solution:
$\sec x+\tan x=\sqrt{3}\cdots \left(1\right)$
and we know that $\sec x-\tan x=\frac{1}{\sec x+\tan x}=\frac{1}{\sqrt{3}}\cdots \left(2\right)$
From $\left(1\right)+\left(2\right)$
$\sec x=\frac{2}{\sqrt{3}}$
From $\left(1\right)-\left(2\right)$
$\tan x=\frac{1}{\sqrt{3}}$
$\tan x,\sec x$ both are positive only in $1^{\text{st}}$ quadrant.
So, $x=\frac{\pi }{6},\frac{13\pi }{6}$
Hence, the number of solutions is $2.$