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 ?

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 $x = 2 \, n \, \pi - \tan^{-1}(\text{some value})$.
