<s>[INST] 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.
 [/INST] Using $\displaystyle \bf{A.M\geq G.M}$
$$\displaystyle \frac{4\sin^2 x+\csc^2 x}{2}\geq \sqrt{4\sin^2 x\cdot \csc^2 x}\Rightarrow 4\sin^2 x+\csc^2 x\geq 4$$
and equality hold when $$4\sin^2 x=\csc^2 x$$
and $$\displaystyle \frac{\tan^2 x+\cot^2 x}{2}\geq \sqrt{\tan^2 x\cdot \cot^2 x}\Rightarrow \tan^2 x+\cot^2 x\geq 2$$
and equality hold when $$\tan^2 x= \cot^2 x$$
Now adding these two , We get $$\displaystyle 4\sin^2 x+\csc^2 x+\tan^2 x+\cot^2 x\geq 6$$
and here equality condition is satisfied
So we get $$4\sin^2 x=\csc^2 x$$ and $$\tan^2 x= \cot^2 x$$ where $0\leq x\leq 2\pi$
</s>