1. ## Equation

Solve $\displaystyle \cos^2x+\cos^22x+\cos^23x=1$

2. ## Cos(n*x)

First, you need to express $\displaystyle \cos(2x)$ and $\displaystyle \cos(3x)$ in terms of $\displaystyle \cos x$. ($\displaystyle \cos (nx)$ is for positive integer n always an nth degree polynomial of $\displaystyle \cos x$; they're called the Chebyshev_polynomials)
For $\displaystyle \cos(2x)$ you should have the double angle identity $\displaystyle \cos(2x) = 2\cos^2 x - 1$.
For $\displaystyle \cos(3x)$, use the cosine sum formula with 3x=2x+x to get:
$\displaystyle \cos(3x)=\cos(2x+x)$
$\displaystyle =\cos(2x)\cos x - \sin(2x)\sin x$
$\displaystyle =(2\cos^2 x - 1)\cos x - 2\sin^2 x \cos x$ (from the double angle identities)
$\displaystyle =2\cos^3 x - \cos x - 2(1-\cos^2 x) \cos x$
$\displaystyle =4\cos^3 x - 3\cos x$

Thus:
$\displaystyle \cos^2 x + \cos^2 2x + \cos^2 3x = 1$
$\displaystyle \cos^2 x + (2\cos^2 x - 1)^2 + (4\cos^3 x - 3\cos x)^2 = 1$
$\displaystyle \cos^2 x + 4\cos^4 x - 4\cos^2 x + 1 + 16\cos^6 x - 24\cos^4 x + 9\cos^2 x= 1$
$\displaystyle 16\cos^6 x - 20\cos^4 x + 6\cos^2 x= 0$ (simplifying, and subtracting one from both sides)
$\displaystyle 8\cos^6 x - 10\cos^4 x + 3\cos^2 x= 0$ (dividing by two)

This is an easily factorable polynomial, and should give you three equations of the form $\displaystyle cos^2 x = a$ for some a, which you can then solve.

--Kevin C.

3. Complex numbers can also solve the equation.

$\displaystyle \cos x = \frac{{e^{ix} + e^{ - ix} }} {2}.$

This one becomes nicer.