Question

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

Answer 1

$\frac{\sqrt{3}-1}{\sin x}+\frac{\sqrt{3}+1}{\cos x}=4\sqrt{2}$

Divide both sides by $2\sqrt{2}$

$\frac{\sqrt{3}-1}{2\sqrt{2}\sin x}+\frac{\sqrt{3}+1}{2\sqrt{2}\cos x}=2$

Multiply both sides by $\sin x\cos x$

$\frac{\sqrt{3}-1}{2\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{2\sqrt{2}}\sin x=2\sin x\cos x$

$\Rightarrow \sin {15}^{\circ }\cos x+\cos {15}^{\circ }\sin x=\sin 2x$ since

$\sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$ and $\cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\Rightarrow \sin ({15}^{\circ }+x)=\sin 2x$

$\Rightarrow {15}^{\circ }+x=2x,{15}^{\circ }+x={180}^{\circ }-2x$

$\Rightarrow {15}^{\circ }=x,x+2x={180}^{\circ }-{15}^{\circ }={165}^{\circ }$

$\Rightarrow x={15}^{\circ },3x={165}^{\circ }$

$\therefore x={15}^{\circ },x=\frac{{165}^{\circ }}{3}={53}^{\circ }$