Question

The general solution of $x$ satisfying ${\sin }^{2}x.\sec x+\sqrt{3}\tan x=0$ is given by

• A. $x=\frac{n\pi }{2}$
• B. $x=\frac{n\pi }{3}$
• C. $x=n\pi$
• D. $x=2n\pi$

Answer 1

The correct option is C $x=n\pi$

${\sin }^{2}x\sec x+\sqrt{3}\tan x=0$

$\Rightarrow \frac{{\sin }^{2}x}{\cos x}+\sqrt{3}\tan x=0$

$\Rightarrow \sin x\tan x+\sqrt{3}\tan x=0$

$\Rightarrow \tan x(\sin x+\sqrt{3})=0$

$\Rightarrow \tan x=0$ or $\sin x+\sqrt{3}=0$

$\Rightarrow \tan x=0$ or $\sin x\ne -\sqrt{3}\because rangeof\sin x\in [-1,1]$

$\therefore \tan x=0$

$\therefore x=n\pi$