Solve the equation, \[\cos 4\theta =\cos 2\theta \].

Hint: At first take \[\cos 4\theta \] from left to right side of equation and then apply identity that is, \[\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)\].
Then equate it to 0. Hence, use the formula if \[\sin x=\sin \alpha \], then \[x=n\pi +{{\left( -1 \right)}^{n}}\alpha \], where n is any integer and \[\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\].

Complete step-by-step answer:

In the question we are given an equation, \[\cos 4\theta =\cos 2\theta \] and we find the general values or solutions for \[\theta \] for which equation satisfy.

We are given equation that,

\[\cos 4\theta =\cos 2\theta \]

So, we can write it as,

\[\cos 2\theta -\cos 4\theta =0\]

We know a formula that,

\[\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)\]

And we can take A as \[2\theta \] and B as \[4\theta \]. So, we can rewrite the equation as, \[2\sin \left( \dfrac{2\theta +4\theta }{2} \right)\sin \left( \dfrac{4\theta -2\theta }{2} \right)=0\]

Or

\[2\sin 3\theta \sin \theta =0\]

So, for \[\sin 3\theta =0\] or \[\sin \theta =0\], the equation satisfies.

So, let’s take \[{{1}^{st}}\] case where, \[\sin 3\theta =0\].

We know that, \[\sin \theta =0\]. So, we can write equation as,

\[\sin 3\theta =\sin 0\]

We can say that, \[3\theta =n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)\] or \[3\theta =n\pi \] or \[\theta =\dfrac{n\pi }{3}\], by using formula, if \[\sin x=\sin \alpha \], then \[x=n\pi +{{\left( -1 \right)}^{n}}\alpha \], where n is any integer and \[\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\].

So, for \[{{2}^{nd}}\] case, where \[\sin x\] is equal to 0.

We can write the equation, \[\sin x=0\], as \[\sin \theta =\sin 0\].

So, we can say that, \[\theta =n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)\] or \[\theta =n\pi \].

Hence, the general solutions are \[n\pi \] and \[\dfrac{n\pi }{3}\].

Note: One can also solve by converting \[\cos 4\theta \] in terms of \[\cos 2\theta \] by using relation that, \[\cos 4\theta =2{{\cos }^{2}}2\theta -1\] and then factorize it but that will be a bit confusing process.