Question

The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2x|,x\in (-\frac{\pi }{2},\frac{\pi }{2})-\{\frac{\pi }{4},-\frac{\pi }{4}\}$ is

• A. $\frac{\pi }{10}$
• B. $-\frac{7\pi }{30}$
• C. $-\frac{11\pi }{30}$
• D. $-\frac{\pi }{15}$

Answer 1

The correct option is C $-\frac{11\pi }{30}$

$\frac{\cos x}{1+\sin x}=|\tan 2x|,x\in (\frac{\pi }{2},\frac{\pi }{2})-\{\frac{\pi }{4},-\frac{\pi }{4}\}$
If $x\in (0,\frac{\pi }{4})$, then $\frac{\cos x}{1+\sin x}=\frac{2\tan x}{1-{\tan }^{2}x}$
$\Rightarrow \frac{\cos x}{1+\sin x}=\frac{2\sin x\cdot \cos x}{{\cos }^{2}x-{\sin }^{2}x}$
$\Rightarrow {\cos }^{2}x-{\sin }^{2}x=2\sin x+2{\sin }^{2}x(\because \cos x\ne 0)$
$\Rightarrow 4{\sin }^{2}x+2\sin x-1=0$
$\therefore \sin x=\frac{\sqrt{5}-1}{4},\frac{-\sqrt{5}-1}{4}\text{(Not applicable)}$
$\therefore x=\frac{\pi }{10}\cdots \left(i\right)$

If $x\in (-\frac{\pi }{2},-\frac{\pi }{4}),$ then again
$\sin x=\frac{\sqrt{5}-1}{4},-\left(\frac{\sqrt{5}+1}{4}\right)$
$\therefore x=-\frac{3\pi }{10}\cdots \left(ii\right)$

If $x\in (\frac{\pi }{4},\frac{\pi }{2})\cup (-\frac{\pi }{4},0),$ then
$1-2{\sin }^{2}x=-2\sin x-2{\sin }^{2}x$
$\therefore \sin x=-\frac{1}{2}$
$\therefore x=-\frac{\pi }{6}\cdots \left(iii\right)$
$\therefore$ Sum of solution $=\frac{\pi }{10}-\frac{3\pi }{10}-\frac{\pi }{6}=-\frac{11\pi }{30}$