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.
$\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) $.
No, the quadratic equation is already simplified as you gave, which can be solved using numerical methods like Newton Raphson or any other standard numerical equation method. There are no more straightforward ways to solve the equation; you already used the standard trigonometric way, which is to convert cos(x) and sin(x) into the same trigonometric ratio. Although the solution can be analyzed using a graph.
You can use the explicit formula for the roots of an equation of degree four. See this post, for example:
Is there a general formula for solving 4th degree equations (quartic)?
If you input the equation into Mathematica or Wolfram Alpha, it will give you explicit complicated formulas such as those.
Your solution is fine but you must take care that squaring introduce extra solutions.
For $x \in (0,2\pi)$, the original equation has two roots but after squaring two other appear.
In any manner, solving the quartic is possible but the result is awful (ask for the exact form).
Now, almost for the fun
By inspection, the solution is close to $x=\frac \pi 3$. What you can do is to expand $$f(x)=\cos(x)-\sin^4(x)$$ around this value to have $$f(x)=\sum_{n=0}^\infty a_n\,\left(x-\frac{\pi }{3}\right)^n$$ where the first coefficients are $$\left\{-\frac{1}{16},-\frac{5 \sqrt{3}}{4},-\frac{1}{4},\frac{13}{4 \sqrt{3}},\frac{25}{48},-\frac{29}{16 \sqrt{3}},-\frac{481}{1440},\frac{2113}{3360 \sqrt{3}},\frac{1613}{16128},-\frac{6605}{48384 \sqrt{3}},\cdots\right\}$$ Truncate to some order and then use series reversion to get $$x=\frac \pi 3-\sum_{n=1}^p b_n \,t^n\qquad \text{where} \qquad t=\frac{4 }{5 \sqrt{3}}\left(f(x)+\frac{1}{16}\right)$$ and we want $f(x)=0$.
The first $b_n$ are $$\left\{1,\frac{1}{5 \sqrt{3}},\frac{67}{75},\frac{139}{300 \sqrt{3}},\frac{41591}{22500},\frac{294541}{225000 \sqrt{3}},\frac{38163739}{7875000},\frac{182178917}{45000000 \sqrt{3}},\frac{1921145413}{135000000},\cdots\right\}$$ Using them, an approximation is $$x \sim \frac{\pi }{3}-\frac{1967557778378954379031}{39191040000000000000000 \sqrt{3}}=\color{red}{1.01821209908832}93$$ while the "exact" solution is $$x=\color{red}{1.0182120990883258}$$
Edit
Using twice more terms in the expansion, $$x \sim \frac{\pi }{3}-\frac{2295379042419019204745481645746329171766630434241}{45720787900170240000000000000000000000000000000000 \sqrt{3}}$$ showing an absolute error equal to $3.47\times 10^{-26}$.
