Is it possible to solve analytically $x^4+x^2+\cos x=0$

Question

For the equation $x^4+x^2+\cos x=0$, it is possible to solve it analytically? If not, which numerical method can I employ to find its root(s)?

Thank you!

Answer 1

Use Newton method.

Being lazy, I started using $x_0=1+i$. This generates the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.0000000000+1.0000000000\, i \\ 1 & 0.7689041871+0.8420052018\, i \\ 2 & 0.6431803621+0.7829949990\, i \\ 3 & 0.6093767798+0.7799884053\, i \\ 4 & 0.6078614041+0.7810154976\, i \\ 5 & 0.6078632528+0.7810203081\, i \end{array} \right)$$ which is the solution for ten significant figures.

Since the function is even, you have the second root by symmetry.

Edit

If you enjoy nightmares, let $x=a +i b$, replace and expand. This would lead you to two "nice" equations $$a^4-6 a^2 b^2+a^2+b^4-b^2+\cos (a) \cosh (b)=0$$ $$2 a b \left(2 a^2-2 b^2+1\right)-\sin (a) \sinh (b)=0$$ corresponding to the real and imaginary parts.

You could solve them for $(a,b)$ using Newton-Raphson method.

Answer 2

John Duma showed that there are no real roots.

Wolfy says that it has complex roots $x ≈ \pm 0.607863 \pm 0.78102 i$.

Answer 3

The elementary functions are generated from their variable by applying finite numbers of algebraic functions (algebric operations), $\exp$ and/or $\ln$.

The theorem of Lin 1983 implies that your equation cannot be solved by elementary numbers. That means, the equation cannot be solved by applying only elementary functions.

Your equation cannot be solved by Lambert W either.