<s>[INST] Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it  by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
 [/INST] Use the cosine of sums formula to decompose $\cos 3x$.
$$\cos 3x = \cos (2x + x) = \cos 2x \cdot \cos x - \sin x \cdot \sin 2x$$
The identities $\cos 2x = 2 \cos^2 x - 1$ and $\sin 2x = 2 \sin x \cos x$ are widely known. These are also derived from the cosine of sums and sine of sums formulas respectively. Plugging them into our last result, we have
\begin{align}
\cos 3x &= \left(2 \cos^2 x - 1\right) \cos x - \sin x \cdot 2 \sin x \cos x \\
&= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\
\end{align}
Since $\sin^2 x = 1 - \cos^2 x$, we have
\begin{align}
\cos 3x &= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\
&= 2\cos^3 x - \cos x - 2 \left(1 - \cos^2 x\right) \cos x \\
&= 2\cos^3 x - \cos x - 2 \cos x + 2 \cos^3 x \\
&= 4\cos^3 x - 3 \cos x
\end{align}
At this point, we have eliminated the $\sin$'s and $\cos 3x$ is expressed entirely in terms of a polynomial of $\cos x$. Plugging this into our equation $\cos^3 x + \cos x = 0$, we have
\begin{align}
\left(4\cos^3 x - 3 \cos x\right) + \cos x &= 0 \\
4\cos^3 x - 2 \cos x &= 0 \\
2\cos^3 x - \cos x &= 0
\end{align}
Taking $y = \cos x$, this reduces to the polynomial
$$2y^3 - y = 0 \Rightarrow y(2y^2 - 1) = 0 \Rightarrow y(y - \sqrt{2}/2)(y + \sqrt{2}/2) = 0$$
which gives three distict roots $y = 0$, $y = -\sqrt{2}/2$ and $y = \sqrt{2}/2$. Then, we plug-in $y = \cos x$ into each and solve them separately.
For $y = 0$, we have $\cos x = 0$. This holds for $x = \pi/2$ plus any integer multiple of $\pi$. These are exactly the points where $\cos x$ intersects the $x$-axis.
For $y = -\sqrt{2}/2$, we have $\cos x = -\sqrt{2}/2$. From here, we have exactly the bottom points of the cosine curve $\pm \pi/4$.
For $y = +\sqrt{2}/2$, we have $\cos x = +\sqrt{2}/2$. From here, we have exactly the peak points of the cosine curve $\pm \pi/4$.
Taking the union of these solution sets, we get the solution set for $\cos^3 x + \cos x = 0$ which are exactly the multiples of $\pi/4$ that are not multiples of $\pi/2$.
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