The smallest value of x satisfying the equation
Question:

The smallest value of $x$ satisfying the equation $\sqrt{3}(\cot x+\tan x)=4$ is

(a) $2 \pi / 3$

(b) $\pi / 3$

(c) $\pi / 6$

(d) $\pi / 12$

Solution:

(c) $\pi / 6$

Given:

$\sqrt{3}(\cot x+\tan x)=4$

$\Rightarrow \sqrt{3}\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)=4$

$\Rightarrow \sqrt{3}\left(\cos ^{2} x+\sin ^{2} x\right)=4 \sin x \cos x$

$\Rightarrow \sqrt{3}=2 \sin 2 x \quad[\sin 2 x=2 \sin x \cos x]$

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

$\Rightarrow \sin 2 x=\sin \frac{\pi}{3}$

$\Rightarrow 2 x=n \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}, n \in Z$

$\Rightarrow x=\frac{n \pi}{2}+(-1)^{n} \frac{\pi}{6}, n \in Z$

To obtain the smallest value of $x$, we will put $n=0$ in the above equation.

Thus, we have:

$x=\frac{\pi}{6}$

Hence, the smallest value of $x$ is $\frac{\pi}{6}$.