Question

General solution of the equation $2{\sin }^{2}x+3{\cot }^{2}x-4\sin x-6\cot x+5=0$ is

• A. $x=(4n+1)\frac{\pi }{2}\text{where}n\in Z$
• B. $x=(4n+1)\frac{\pi }{4}\text{where}n\in Z$
• C. $x=(2n+1)\frac{\pi }{2}\text{where}n\in Z$
• D. $x=\varphi$

Answer 1

The correct option is D $x=\varphi$

$2{\sin }^{2}x+3{\cot }^{2}x-4\sin x-6\cot x+5=0$
$\Rightarrow 2{\sin }^{2}x-4\sin x+2+3{\cot }^{2}x-6\cot x+3=0$
$\Rightarrow 2(\sin x-1{)}^2+3(\cot x-1{)}^2=0$
$\Rightarrow \sin x-1=0$ and $\cot x-1=0$
$\Rightarrow \sin x=1$ and $\cot x=1$
But when $\sin x=1\Rightarrow \cot x=0$ as $\cot x=\frac{\cos x}{\sin x}$
So, $x=\varphi$