I am looking for the solutions of the trigonometric equation
$$\cos(x) =\sin^4(x)$$
I tried this way:
$$\begin{align}
\cos(x) =\sin^4(x) &\iff \cos^2(x) =\sin^8(x) \\
&\iff 1-\sin^2(x) -\sin^8(x)=0 \\
&\iff 1-u-u^4=0
\end{align}$$
if I put $u=\sin^2(x)$.
Does this equation have solutions which can be explicitly written? I mean, no numerical methods or something like that.
There is an easier way to solve it (better than mine)?
Thank you.
Best Answer
$\cos(x) = \sin^4 (x) = (\sin^2(x))^2 = (1 - \cos^2(x))^2 = 1 - 2 \cos^2(x) + \cos^4(x) $
Letting $u = \cos(x)$ , we now have the following quartic polynomial equation:
$u^4 - 2 u^2 - u + 1 = 0$
Being a quartic polynomial equation, it does have closed form solutions. For more details on how to find the roots of a quartic polynomial function, check this page. Since this is obviously complicated, you may resort to numerical methods such as the bisection method, or Newton's method (which is much faster). Once you find all the roots $u$ of the above polynomial, then select the ones that are real and having a magnitude that lies in $u \in [-1, 1]$, then the corresponding angle is $ x = \cos^{-1}(u) $.