Question

The general value of $x$ satisfying the equation $2{\cot }^{2}x+2\sqrt{3}\cot x+4cosecx+8=0$

• A. $n\pi -\frac{\pi }{6},nϵZ$
• B. $n\pi +\frac{\pi }{6},nϵZ$
• C. $2n\pi -\frac{\pi }{6},nϵZ$
• D. $2n\pi +\frac{\pi }{6},nϵZ$

Answer 1

The correct option is C $2n\pi -\frac{\pi }{6},nϵZ$

$2{\cot }^{2}x+2\sqrt{3}\cot x+4\csc x+8=0$
$\Rightarrow 2{\cos }^{2}x+2\sqrt{3}\cos x\sin x+4\sin x+8{\sin }^{2}x=0$
$\Rightarrow {(\sqrt{3}\sin x+\cos x)}^2+4{\sin }^{2}x+4\sin x+1=0$
$\Rightarrow {(\sqrt{3}\sin x+\cos x)}^2+{(2\sin x+1)}^2=0$
$\Rightarrow \tan x=\frac{-1}{\sqrt{3}}$ and $\sin x=\frac{-1}{2}$
i.e $x=-\frac{\pi }{6}$
Hence, general solution is $2n\pi -\frac{\pi }{6},nϵZ$