<s>[INST] 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 $x$?

 [/INST] Yet an other way to proceed, using algebraic only manipulations after a quick reshape...
We want to show
$$
\frac 58\cdot \frac{\cos 36^\circ}{\sin 36^\circ}
\overset{!}=\sin^3 72^\circ \ .
$$
Let us denote by $c,s$ respectively the values for $\cos 36^\circ$ and $\sin 36^\circ$. Then we want:
$$
\frac 58\cdot\frac cs 
\overset{!}=
8s^3c^3\ ,\text{ or equivalently }
5 - 64s^4c^2=0\ .
$$
We start with $(s+ic)^5=\cos (5\cdot 36^\circ)+i\sin(5\cdot 36^\circ)=-1$, and consider only the imaginary part, thus getting
$$ 
\begin{aligned}
0 
&= 5sc^4-10s^3c^2+s^5\ ,\text{ and since $s\ne 0$ we get}\\
0 &= 5c^4-10s^2c^2+s^4\\
&=5(1-2s^2+s^4)-10s^2(1-s^2)+s^4\\
&=5-20s^2+16s^4\ .
\qquad\text{From here:}
\\[3mm]
5-64s^4c^2
&=5 - 64s^4(1-s^2)
\\&=
5-64s^4+64s^6\\
&=\underbrace{(5-20s^2+16s^4)}_{=0}(1+4s^2)=0\ .
\end{aligned}
$$
$\square$
</s>