I have to find the general solution of this trigonometric equation
$$\cos4 \theta = \cos2 \theta $$
I solved in the following manner, but I got the wrong answer.
$$\begin{align}
\\\cos4 \theta &= \cos2 \theta \\
2\cos^2 2\theta -1&=\cos2 \theta \\
(\cos2 \theta-1)(2\cos2 \theta +1)&= 0
\end{align}$$
$$
\cos2 \theta= 1 \text{ or }-\frac 12 \\[12pt]
\theta = n\pi \text{ or }n\pi \pm \frac\pi3
$$
Can anyone tell me what's I'm missing? Thanks in advance.
Best Answer
From your solution the values of $3\theta$ are $$3n\pi,(3n\pm1)\pi$$
Now using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$0=\cos2\theta-\cos4\theta=2\sin\theta\sin3\theta$$
So, it is sufficient to have $$\sin3\theta=0\implies3\theta=m\pi$$ where $m$ is an integer
Observe that $m$ can be represented by at least one of the forms $$3n,3n-1,3n+1$$
So, the two results actually coincide