Finding all the solutions of $\cos2\theta -\cos4\theta =0$ $$\cos2\theta -\cos4\theta =0 $$  in the interval $[0,2\pi)$
So I know that I can use the Double or Half Angle formulas to simplify this and help me find the solutions. So I used the Double Angle Formulas and got:
$$2\cos^{ 2 }\theta -1-2\cos^{ 2 }2\theta -1=0$$
Now things get hairy with this value and I don't know how to go about solving from here.
$$2\cos^{ 2 }2\theta -1$$
 A: Use half angle formula only for $\cos 4\theta$ so that you get a quadratic equation in $\cos 2\theta$.
$$2\cos^2 2\theta - \cos 2\theta -1 = 0$$
Solve the equation for $\cos 2\theta$ and then for $\theta$.
A: Hint:
$$0=\cos2\theta-\cos4\theta=\cos2\theta-2\cos^22\theta+1$$
Now solve the almost trivial quadratic.
A: Whenever you have $\cos A=\cos B,$ we can safely say $A=2n\pi\pm B$  where $n$ is any integer
$$\cos4\theta=\cos2\theta \implies4\theta=2m\pi\pm2\theta$$ where $m$ is any integer
Taking the '-' sign, $\theta=\dfrac{m\pi}3\  \ \ \ (1)$
Taking the '+' sign, $\theta=m\pi$ which is a proper subset of $(1)$
Now we need $0\le \dfrac{m\pi}3<2\pi\iff 0\le m<6$
A: $cos4\theta=2cos^22\theta-1=2(2cos\theta-1)^2-1=2(4cos^2\theta+1-4cos\theta)-1$
The original equation will then become
$8cos^2\theta+2-8cos\theta-1-2cos^2\theta+1=0$
$6cos^2\theta-8cos\theta+2=0$
$6cos\theta(cos\theta-1)-2(cos\theta-1)=0$
So,
$cos\theta=\frac{1}{3}$ 
Or,
$cos\theta=\frac{1}{2}$
:)
A: $$\cos^2{\theta}-\sin^2\theta=2\cos2\theta-1$$
$$\cos^2\theta-\sin^2\theta=2\cos^2\theta-2\sin^2\theta -1$$
$$3\sin^2\theta=\cos^2\theta-1$$
$$3\sin^2\theta=sin^2\theta$$
$$2\sin^2\theta=0$$
or $\sin^2\theta=0$
$$\implies \theta=0$$
Therefore general solution is $\theta=n\pi$
