Question

The smallest value of x satisfying the equation $\sqrt{3}\left(\cot x+\tan x\right)=4$ is
(a) $2\pi /3$
(b) $\pi /3$
(c) $\pi /6$
(d) $\pi /12$

Answer 1

(c) $\pi /6$

Given:

$$\begin{gathered} \sqrt{3}(\cot x+\tan x)=4 \\ \Rightarrow \sqrt{3}\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)=4 \\ \Rightarrow \sqrt{3}({\cos }^{2}x+{\sin }^{2}x)=4\sin x\cos x \\ \Rightarrow \sqrt{3}=2\sin 2x[\sin 2x=2\sin x\cos x] \\ \Rightarrow \sin 2x=\frac{\sqrt{3}}{2} \\ \Rightarrow \sin 2x=\sin \frac{\mathrm{\pi }}{3} \\ \Rightarrow 2x=n\mathrm{\pi }+(-1{)}^{\mathrm{n}}\frac{\mathrm{\pi }}{3},n\in Z \\ \Rightarrow x=\frac{n\mathrm{\pi }}{2}+(-1{)}^n\frac{\mathrm{\pi }}{6},n\in Z \end{gathered}$$
To obtain the smallest value of x, we will put $n=0$ in the above equation.
Thus, we have:
$x=\frac{\mathrm{\pi }}{6}$
Hence, the smallest value of x is $\frac{\mathrm{\pi }}{6}$.