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.
 A: Other than the cosmetic aspect of choosing the exponents in an arithmetic progression, it comes to the fact that there is a nice clean formula for the maximum value of $|\sin^a (x) \cos^b (x)|$, because that function is maximized when $\sin^2 (x) = \frac{a}{a+b}$ (for $a,b > 0$).  Then the problem is set by asking for the $x$ where the function is equal to a value that just happens to equal the nice formula for the maximum.
The location of the maxima can be proved by calculus or (as hinted in the question) with the Arithmetic Mean-Geometric Mean inequality applied to $a(\frac{\sin^2}{a}) + b(\frac{\cos^2}{b}) = 1$.  
The 3:1 ratio of $a$ and $b$ seen in this set of exponents is the only nontrivial case for which $x$ is a rational multiple of $\pi$.
A: 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.
A: So after reducing the equation the equation will look like that:
$$27\csc^6(x)\sec^2(x)=256$$
OR:
$$\sin^6(x)\cos^2(x)=\frac{27}{256}$$
Then:
$$\sin^6(x)-\sin^8(x)=\frac{27}{256}$$
Or:
$$\sin^8(x)-\sin^6(x)+\frac{27}{256}=0$$
I found that $\sin^2(x)-\frac{3}{4}=0$ is divisible above equation.I there used the features of high degree polynomials.
So the equation will look like in this way:
$$(\sin^2(x)-\frac{3}{4})(\sin^6(x)+\frac{1}{4}\sin^4(x)+\frac{3}{4}\sin^2(x)+\frac{27}{64})=0$$
And:
$$\sin^2(x)-\frac{3}{4}=0$$ 
The rest is for you.
A: 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.
A: 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
