The original question :
Find all $x$ in $\mathbb R$ such that $16\sin^3(x) -14\cos^3(x) = \sqrt[3]{\sin x\cos^8(x) + 7\cos^9(x)}$
It's a tough question I've found. I've tried using
$16\tan^3(x) -14 = \sqrt[3]{\tan x + 7 }$
By inspection, $\tan x=1$ is one of the answers.
but According to WA, $\tan x$ is not equal to $1$.
(Sorry , I've seen later that $x = \frac{\pi}{4}$ , but I don't know how to find all roots of the question.)
Can root of unity solve this?
The function $f\colon \mathbb{R} \to \mathbb{R}$, $f(a) = (2a)^3-7$, is bijective and increasing. Since $f(a) - a = (a-1)(8a^2+8a+7)$, the only solution of the equation $f(a) = a$ is $a=1$; $f(a) > a$ if $a>1$ and $f(a)<a$ if $a<1$. Then the equations $f(f(a)) = a$ and $f(a) = a$ are equivalent. In particular, $f^{-1} (a) = f(a)$ if and only if $a=1$. In the original problem, $a=\tan{x}$.