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.

There is actually quite an ingenious solution to this question. We have $\sin^6(x)\cos^2(x) = \dfrac{27}{256}$. Now we can write this as $3^3\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\cos^2(x) = \dfrac{27}{256}$. Then by AM-GM $\sqrt[4]{\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\cos^2(x)} \leq \dfrac{1}{4} $. Thus, $3^3\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\dfrac{\sin^2(x)}{3}\cos^2(x) \leq \dfrac{27}{256}$. Therefore equality holds iff $\sin^2(x) = 3\cos^2(x) \implies 3-4\sin^2(x) = 0 \implies \sin^2(x) = \dfrac{3}{4}$ and the solution proceeds.
