# 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?

• "Another way"? You didn't even show the way you tried to do it. Jun 22, 2020 at 2:15

So, we have $${\cos(2x + x) + \cos(x) = 0}$$. Using the compound trig formulae:

$${\Rightarrow \cos(2x)\cos(x) - \sin(2x)\sin(x) + \cos(x) = 0}$$

Using the double angle identity for $${\sin(2x)}$$, and $${\cos(2x)}$$ gives

$${(2\cos^2(x) - 1)\cos(x) - 2\cos(x)\sin^2(x) + \cos(x)= 0}$$

$${\Rightarrow 2\cos^3(x) - 2\cos(x)\left(1 - \cos^2(x)\right)=0}$$

$${\Rightarrow 4\cos^3(x) - 2\cos(x) = 0}$$

$${\Rightarrow \cos(x)\left(4\cos^2(x) - 2\right)=0}$$

So either $${\cos(x)=0}$$ or $${4\cos^2(x) - 2 = 0}$$. Can you take it from here?

I think it is easier to use addition formula $$\cos(a)+\cos(b)=2\cos(\frac{a+b}2)\cos(\frac{a-b}2)$$ in this case rather than expanding $$\cos(nx)$$.

You get directly the $$\frac \pi 2+k\pi$$ and $$\frac \pi 4+\frac k2\pi\$$ lines from the product:

$$2\cos(x)\cos(2x)=0$$

$$\cos(3x)+ \cos(x)=0$$ $$\cos(3x)=- \cos(x)$$ $$\cos(3x)= \cos(\pi-x)$$ $$3x= 2k\pi\pm (\pi-x)$$ $$x=\frac{(2k+1)\pi}{4}, \ \ \frac{(2k-1)\pi}{2}$$ $$x\in\{\frac{(2k+1)\pi}{4}\}\cup\{\frac{(2k-1)\pi}{2}\}$$ Where, $$k$$ is an integer i.e. $$k=0, \pm1, \pm2, \ldots$$

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 $$\pi/2$$.

Yet another method using

$$cos(3x)=4cos^3(x)-3cos(x)$$

$$cos(3x)-cos(x)=0$$

$$4cos^3(x)=4cos(x)$$

$$cos^3(x)=cos(x)$$

$$cos(x)=0$$ ----(1)

Or

$$cos^2(x)=1$$

$$sin^2(x)=0$$ -----(2)

Use (1) and (2) to find general solutions

Symmetry is always a powerful and nice instrument \eqalign{ & 0 = \cos (3x) + \cos x = \cr & = \cos (2x + x) + \cos (2x - x) = \cr & = \cos (2x)\cos x - \sin (2x)\sin x + \cos (2x)\cos x + \sin (2x)\sin x = \cr & = 2\cos (2x)\cos x \cr}

and then \eqalign{ & \cos (2x) = 0\quad \vee \quad \cos x = 0 \cr & 2x = {\pi \over 2} + k\pi \quad \vee \quad x = {\pi \over 2} + j\pi \cr & x = {\pi \over 4} + k{\pi \over 2}\quad \vee \quad x = {\pi \over 2} + j\pi \cr & x = {\pi \over 4} + 2k{\pi \over 4}\quad \vee \quad x = 2{\pi \over 4} + 4j{\pi \over 4} \cr & x = \left( {2k + 1} \right){\pi \over 4}\quad \vee \quad \left( {2j + 1} \right){\pi \over 2}\quad \left| {\,k,j \in Z} \right. \cr}

• Your last line is wrong, you can reach $a\frac{\pi}4+2n\pi$ for $a=1,2,3,5,6,7$ so you are missing $2,6$ in your formula. i.e. for $j=0$ then $x=\frac{\pi}2$ which is not in your final formula.
– zwim
Jun 25, 2020 at 17:47
• @zwim oops, banal mistake ! thanks for signalling ! Jun 25, 2020 at 21:54