Solve the equation: $\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$ How to solve the equation 

$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$

Can anyone give me some hints?
 A: HINT:
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1\Longleftrightarrow$$
$$1+6\cos^2(x)-20\cos^4(x)+16\cos^6(x)=1\Longleftrightarrow$$

Substitute $y=\cos^2(x)$:

$$16y^3-20y^2+6y=0\Longleftrightarrow$$
$$2y(2y-1)(4y-3)=0\Longleftrightarrow$$
$$y(2y-1)(4y-3)=0$$
A: Hint: Simplify $1+6\cos^2(x)-20\cos^4(x)+16\cos^6(x)=1$ and then substitute $y=\cos^2(x)$ and then you will get $16y^3-20y^2+6y+1=1$
A: HINT: $$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
$$\cos^2(x)+\cos^2(3x)=1-\cos^2(2x)$$
$$\cos^2(x)+\cos^2(3x)=(1-\cos(2x))(1+\cos(2x))$$
$$\cos^2(x)+\cos^2(3x)=(2\sin^2(x))(2\cos^2(x))$$
$$\cos^2(x)+(4\cos^3(x)-3\cos (x))^2=2(1-\cos^2(x))(2\cos^2(x))$$
$$16\cos^6x-20\cos^4x+6\cos^2x=0$$
now, let $\cos^2 x=t$, $$8t^2-10t+3t=0$$
$$t(2t-1)(4t-3)=0$$
A: Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)-1$$
$$=\cos^2(x)-\sin^23x+\cos^2(2x)$$
$$=\cos(3x-x)\cos(3x+x)+\cos^2(2x)$$
$$=\cos2x(\cos4x+\cos2x)$$
Now use Prosthaphaeresis Formulas on $$\cos4x+\cos2x$$
Should I use a single word more?

Alternatively using $\cos2A=2\cos^2A-1,$
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
$$\iff\cos2x+\cos4x+\cos6x+1=0$$
$$\cos6x=2\cos^23x-1$$ and use Prosthaphaeresis Formula on $$\cos2x+\cos4x$$
A: $\cos^2(x)+[2 \cos^2x-1]^2 +[4 \cos^3 x-3 \cos x]^2 = 1$, Simplify the equation to get: $ \cos^2 x [8 \cos^4 x - 10 \cos^2 x + 3] = 0$. Hope this will help.
