Question

Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$

• A. $x=n\pi +\frac{\pi }{6}$,, $n\in Z$
• B. $x=\frac{n\pi }{2}-\frac{\pi }{6}$ , $n\in Z$
• C. $x=n\pi +\frac{\pi }{3}$,, $n\in Z$
• D. $x=\frac{n\pi }{2}-\frac{\pi }{12}$ , $n\in Z$

Answer 1

Answer: (A), (D)

$x=n\pi +\frac{\pi }{6}$,, $n\in Z$
$x=\frac{n\pi }{2}-\frac{\pi }{12}$ , $n\in Z$

$8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$

$8{\sin }^{2}x\cos x=\sqrt{3}\sin x+\cos x$

$4\left(2\sin x\cos x\right)\sin x=\sqrt{3}\sin x+\cos x$

$2\left(2\sin 2x\sin x\right)=\sqrt{3}\sin x+\cos x$

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

$\cos x-\sqrt{3}\sin x=3\cos 3x$

$\cos 3x=\cos (x+\frac{\pi }{3})$

$\Rightarrow$ $3x=2n\pi \pm (x+\frac{\pi }{3})$, $n\in Z$

By taking positive sign, $x=n\pi +\frac{\pi }{6}$, $n\in Z$

By taking negative sign, $x=\frac{n\pi }{2}-\frac{\pi }{12}$, $n\in Z$