How to solve goniometric equation where function arguments differ?

Question

I'm preparing myself for exams and I stumbled over goniometric equation:

$$\cos^2x + \cos{2x}+1=0$$

Normally, those equations are solved by pretending that $\cos{x}$ is some variable $u$ and ending up with something like (in case of quadratic equation):

$$\cos{x} = a_1\\ \cos{x} = a_2$$

But since I cannot substitute $u$ for $\cos{x}$ what should I do? I had a wild idea of removing cosinus the way you can "remove" logarithms. But I don't think this is valid:

$$\cos^2x + \cos{2x}+1=0\\ \cos^2x + \cos{2x}+\cos{0}=0\\ x^2+2x+0 = 0 $$ Even if that was actually valid, what would I do with results for $x$? Can anyone explain the correct approach to this problem?

Answer 1

$$\cos^2x+\cos2x+1=0$$

$$\cos^2x+(2\cos^2x-1)+1=0$$

$$3\cos^2x=0$$

$$\cos x=0$$

$$x=\frac{\pi}{2},\;\frac{3\pi}{2}...$$

Answer 2

An alternative way, using almost the same trigonometric relation that Lanier Freeman applied:

$$\cos^2x+\cos2x+1=0$$

$$\cos^2x+(\cos^2(x)-\sin^2(x))+(\cos^2(x)+\sin^2(x))=0$$

$$3\cos^2x=0$$