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

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!

Solution 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.