Solving $\frac{5}{8} \cot36^\circ = \cos^3x$ without substituting the trig values for $36^\circ$ 
Find the value of $x$ such that 
  $$\frac{5}{8} \cot36^\circ = \cos^3x$$
  The answer is $x=18^\circ$.

It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable value of $x$ and prove that our guess is correct.

Is there any nice way to find the value of $x$?


$$\dfrac58\cdot\cot36^\circ=\dfrac{5\cos36^\circ}{8\sin36^\circ}=\dfrac{5\cos^236^\circ}{4\cos18^\circ}$$
Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
$$\cos36^\circ-(2\cos^236^\circ-1)=\frac12\iff5\cos^2 36^\circ=(1+\cos36^\circ)^2=(2\cos^218^\circ)^2$$
Can you take it from here?
