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.
 A: Hint: with $t=\sin^2x$, 
$$4t^3-12t^2+9t-2 = (2t-1)^2(t-2)$$
A: You can safely assume that all four terms are nonzero, otherwise another one would not be defined. Hence they are all $>0$.
For $t>0$, you have $t+1/t\geq 2$ because $t+1/t-2=(\sqrt t-1/\sqrt t)^2\geq 0$. Hence $\tan^2 x+\cot^2 x\geq 2$.
Likewise, $4t+1/t\geq 4$, because $4t+1/t-4=(2\sqrt t - 1/\sqrt t)^2\geq 0$. Hence $4\sin^2 x+\csc^2 x \geq 4$.
Therefore, you have equality when $|\tan x| = |\cot x|$ and $2|\sin x|=|\csc x|$ (when the expressions in the parentheses above are zero), i.e. when $\tan^2 x=1$ and $\sin^2 x=1/2$, that is when $x=\pi/4+k\pi/2$. There are four such values in $[0,2\pi]$. Notice that the two conditions are equivalent, as $\tan^2 x=1 \iff \sin^2x=1-\sin^2 x \iff \sin^2x=1/2$.
A: 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.
A: 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 
$$x=\frac{(2k+1)\pi}{4} \mbox{   , for } k=0,1,2,3.$$
A: 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 $0\leq x\leq 2\pi$
