# How to solve $\cos^{40}x-\sin^{40}x=1$ for real and imaginary solutions

Solve : $$\cos^{40}x-\sin^{40}x=1$$ I found by induction that $$\sin(x)=0$$ satisfy the equation so $$x=n\pi$$ must be the solution but there should be more solutions real or imaginary , how to find them ?

• Maybe better to say "real or complex". Imaginary might be interpreted as purely imaginary rather than a general complex number. Nov 23, 2022 at 16:25
• Depends what Sushil meant. But if (s)he meant "real or complex", better just say "complex". Nov 23, 2022 at 16:27
• OP, please clarify whether you meant "purely imaginary" i.e. of the form $0 + iy$ and just not purely real i.e. $x + iy$ with $y \neq 0$. Imaginary is most commonly used for the first of these meanings. Nov 23, 2022 at 16:59

HINT

I would recommend you to notice that

\begin{align*} \cos^{40}(x) - \sin^{40}(x) = 1 & \Longleftrightarrow \cos^{40}(x) = \sin^{40}(x) + 1 \geq 1 \end{align*}

Based on such relation, can you proceed from here?

• but it would give only real solutions , what's about imaginary solutions? Nov 23, 2022 at 16:23
• @Sushil if $x$ is imaginary, $\cos^2x$ and $\sin^2x$ are real. Nov 23, 2022 at 16:25

Notice that, with $$t = \cos x$$ and $$y = \sin x$$, the equation factors into the form \begin{align} 1 &= (t - y) (t + y) (t^2 + y^2) (t^4 + y^4) (t^4 - t^3 y + t^2 y^2 - t y^3 + y^4) \\ & \hspace{5mm} \cdot (t^4 + t^3 y + t^2 y^2 + t y^3 + y^4) (t^8 - t^6 y^2 + t^4 y^4 - t^2 y^6 + y^8) \\ & \hspace{5mm} \cdot (t^{16} - t^{12} y^4 + t^8 y^8 - t^4 y^{12} + y^{16}). \end{align} By using $$t^2 + y^2 = 1$$ then this reduces by a factor. There are seven remaining factors. Each polynomial has roots and finding these roots will help lead to the corresponding $$x$$ values. Otherwise it can be determined that $$x = 2 \, n \, \pi$$, $$n = \pm \, \text{integer}$$, and solutions will be of the form $$x = 2 \, n \, \pi - \tan^{-1}(\text{some value})$$.

• I don't understand "make an argument that each factor is equal to one", I don't find it "otherwise evident" that $x\in2\pi\Bbb Z,$ and I can't see how this gives $x = 2 \, n \, \pi - \tan^{-1}(\text{some value}).$ Nov 23, 2022 at 18:47