Help with this trigonometric equation Here is the equation:
$$\sin^5(x) = \frac {\cos(x)}4$$
The answers should be between: 0° and 360°
I really tried and I dont even think it is possible to solve algebraically, maybe using numerical methods or other kind of black magic but that is very advanced for my grade. I am in high school.

\begin{align} 
\sin^5x &= \frac{\cos x}4
,\\
(\sin^5x)^2 &= \left(\frac{\cos x}4\right)^2
,\\
\sin^{10}x &= \tfrac1{16}{\cos^2 x}
,\\
(\sin^{2}x)^5 &= \tfrac1{16}(1-\sin^2 x)
,
\end{align}
\begin{align} 
16(\sin^{2}x)^5+\sin^2 x-1&=0
,\\
(2\sin^2x-1)
(8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1)
&=0
,\\
\cos2x\,(8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1)
&=0.
\end{align}  
\begin{align} 
8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1&>0\quad \forall x\in\mathbb{R}
,
\end{align}
\begin{align} 
\cos2x&=0
,\\
2x&=\tfrac\pi2+\pi k,\quad k=0,1,\dots
,\\
x&=\tfrac\pi4+\tfrac{\pi k}2,\quad k=0,1,\dots
.
\end{align}  
For $x$ restricted to the interval $[0,2\pi]$,
we have four solutions
\begin{align} 
x_1&=\tfrac\pi4=45^\circ
,\\
x_2&=\tfrac\pi4+\tfrac\pi2=\tfrac{3\pi}4=135^\circ
.\\
x_3&=\tfrac\pi4+\pi=\tfrac{5\pi}4=225^\circ
.\\
x_4&=\tfrac\pi4+\tfrac{3\pi}2=\tfrac{7\pi}4=315^\circ
.\\
\end{align}
Substitution of $x_1,\dots,x_4$ into original equation
shows that only solutions $x_1,x_3$ are valid:
\begin{align} 
\sin^5x_1 &=\tfrac{\sqrt2}8
,\\ 
\frac{\cos x_1}4&=\tfrac{\sqrt2}8
,\\
\sin^5x_2 &=\tfrac{\sqrt2}8
,\\ 
\frac{\cos x_2}4&=-\tfrac{\sqrt2}8
,\\
\sin^5x_3 &=-\tfrac{\sqrt2}8
,\\ 
\frac{\cos x_3}4&=-\tfrac{\sqrt2}8
,\\
\sin^5x_4 &=-\tfrac{\sqrt2}8
,\\ 
\frac{\cos x_4}4&=\tfrac{\sqrt2}8
.
\end{align}
As @Malcolm noted, two extra solutions were
introduced by the squaring the original equation,
that's why the final check of the found roots is essential.
