Question

The value of $x$ in $(0,\frac{\pi }{2})$ satisfying $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$ is

• A. $\frac{\pi }{12}$
• B. $\frac{5\pi }{12}$
• C. $\frac{7\pi }{24}$
• D. $\frac{11\pi }{36}$

Answer 1

The correct option is B $\frac{5\pi }{12}$

We have, $\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$
$\Rightarrow \frac{(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x}{\sin x\cos x}=4\sqrt{2}$
$(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2\sqrt{2}\sin 2x$
$\frac{\sqrt{3}-1}{2\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{2\sqrt{2}}\sin x=\sin 2x$
$\sin (x+\frac{5\pi }{12})=\sin 2x$

General solution is
$x+\frac{5\pi }{12}=n\pi +{(-1)}^{n}2x$
For $n=0,x=\frac{5\pi }{12}$
For $n=1,3x=\frac{7\pi }{12}\Rightarrow x=\frac{7\pi }{36}$