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.
On
Yes it has solutions.
$$\sin^5(x) = \frac {\cos(x)}4 $$
$$4\sin^5(x) = \cos(x)$$
$$4\sin^4(x) \tan (x) =1$$
$$ 4(1-\cos ^2(x))^2 \tan (x) =1$$
$$4\tan x (\frac {\tan ^2x}{1+\tan ^2x})^2=1$$
$$4\tan ^5 x = ( 1+\tan ^2x)^2$$
Note that $$ \tan x =1 \implies x= \pi /4 \text { and } 5\pi/4 $$ are solutions.
On
In the interest of completeness I am posting this solution although I don't think that this is necessarily a high school approach. The approach is based on the idea that equations with trigonometric functions are much more amenable to manipulation if there is only one trigonometric function involved. $$ \sin^5(x) = \frac {\cos(x)}4 \\ 4 = \cos x \csc^5 x\\ 4= \cot x \csc^4 x \\ 4 = \cot x (1+\cot^2 x)^2 $$ Or writing $u = \cot x$, $$ u^5 + 2u^3 + u - 4 =0 $$ If we notice that $1+2+1 = 4$ it becomes easier to guess $u=1$ is a solution. In terms of our original variable this is the solution $\cot x =1$ or $x = 45°$ 0r $225°$.
To see if there are any more solutions we could try to factor the other factor in the polynomial, $u^4+u^3+3u^2+3u+4$. We would find that it has no real roots, but it is probably easier to consider the graphs of $y=\sin^5 x$ and of $y=\frac{\cos x}{4}$. We can easily convince ourselves that there are only two intersections in $[0,360]$, so we have found all the solutions.
On
By letting $x=\arctan u$ the problem boils down to solving
$$ (u-1)(4u^4+3u^3+3u^2+u+1)=0$$
and
$$ 4u^4+3u^3+3u^2+u+1 = \left(2u^2+\frac{3}{4}u\right)^2+\left(u+\frac{1}{2}\right)^2+\frac{23}{16}u^2+\frac{3}{4}\geq \frac{3}{4} $$
is always positive, hence $u=1$ (i.e. $x\in\{\frac{\pi}{4},\frac{5\pi}{4}\}$) is the only solution.
Completing the square is a powerful technique: for instance, the decomposition
$$ 4u^4+3u^3+3u^2+u+1 = \left(2u^2+\frac{3}{4}u\right)^2+\frac{39}{16}\left(u+\frac{8}{39}\right)^2+\frac{35}{39} $$
proves that the LHS is $\geq\frac{35}{39}$: this is pretty close to the actual value of the minimum.
\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.