Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$. 
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.

I was hoping some things would cancel out when I expanded this but nothing. I think using inequalities will help.

since
\begin{align*}27\sin{x}\cos^2{x}\tan^3{x}\cot^4{x}\sec^5{x}\csc^6{x}&=27\sin{x}\cos^2{x}\cdot\dfrac{\sin^3{x}}{\cos^3{x}}\cdot\dfrac{\cos^4{x}}{\sin^4{x}}\cdot\dfrac{1}{\cos^5{x}}\cdot\dfrac{1}{\sin^6{x}}\\
&=\dfrac{27}{\cos^2{x}\sin^6{x}}
\end{align*}
so
$$\sin^6{x}\cos^2{x}=\dfrac{27}{256}$$
but on the other hand , we have
$$t(1-t)^3=\dfrac{27}{256}$$
since 
$$ 256t(1-t)^3-27=-(4t-1)^2(16t^2-40t+27)$$
because $$16t^2-40t+27=(4t-5)^2+2>0$$
$$t=\dfrac{1}{4}$$
then
$$\cos{x}=\dfrac{1}{2}\rm{or}-\dfrac{1}{2}$$
where $t=\sin^2{x}\ge 0$
then you can do it
