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.

Let $t=\sin^2 x$. Then, by your simplified expression,
$$p(t)=4t^3 - 12t^2 + 9t - 2 =0 -----(*)$$
The polynomial $p$ factors as
$$p(t)=(2t-1)^2(t-2).$$
So, the solution to $(*)$ is
$$t=\frac{1}{2} \mbox{ or } t=2.$$
Now, we try to solve
$$\sin ^2x = \frac{1}{2} \mbox{ or } \sin^2 x = 2.$$
Well, the rightmost equation does not have a solution. So, we only have to solve the leftmost equation. Taking square roots,
$$\sin x = \pm\frac{1}{\sqrt{2}}.$$
Restricting $x$ to be in $[0,2\pi]$, we see that 
$$x=\frac{(2k+1)\pi}{4} \mbox{   , for } k=0,1,2,3.$$
