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.

Expanding everything out, we get
$$ \frac{256}{27} = \frac{\sin^{1+3}{x}\cos^{2+4}{x}}{\sin^{4+6}{x}\cos^{3+5}{x}},$$
and then cancelling gives
$$ \frac{2^8}{3^3} = \frac{1}{\sin^6{x}\cos^2{x}} $$
Taking a square root and dividing,
$$ \sin^3{x} \cos{x}=\pm \frac{\sqrt{3}^3}{2^4} $$
There is one fairly obvious solution to this: $x=\pm\pi/3$. It is easy to check that the left-hand side has period $\pi$, so $x=k\pi\pm\pi/3$ are solutions. Are there any more? In fact the answer is no: you can show that the global maxima and minima of $\sin^3{x}\cos{x}$ are precisely these points by differentiating it, and hence, by some simple arguments about how the graph must look, there are no more.
