The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$

I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not solve it further.Please help me.Thanks.

Subtract $6$ from both sides of the original equation.
$$4\sin^2x-4+\csc^2x+\tan^2x-2+\cot^2x=0$$
$$(2\sin x-\csc x)^2+(\tan x-\cot x)^2=0$$
Since the square of a real number is non-negative, the only solution is if both squares are $0$.  So we have $\tan x=\cot x$ and
$$2\sin x=\csc x$$
$$\sin^2x=\frac12$$
$$\left\{\begin{eqnarray}\sin x=\pm\frac{\sqrt2}2\\
\cos x=\pm\frac{\sqrt2}2\end{eqnarray}\right.$$
Now all it takes is to verify which of the $4$ solutions to this equation make tangent and cotangent equal.
