Solving $\cos(3x) = \cos(2x)$

I'm struggling with solving given trigonometric equation

$$\cos(3x) = \cos(2x)$$

Let's take a look at the trigonometric identities we can use:

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

and

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

Plugging into the equation and we have that

$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1$$

$$4\cos^3(x) -3\cos(x) - 2\cos^2(x)+1= 0$$

Recalling $t = \cos (x)$,

$$4t^3-2t^2-3t +1 = 0$$

Which is a cubic equation. Your sincerely helps will be appreciated.

Regards!

• With questions like this, I wouldn't usually answer when several others already have, but sometimes I post an answer because I think the others are more complicated than they could be. – Michael Hardy Jun 20 '18 at 16:11
• @MichaelHardy Give different answer to any OP is always a good and very useful habit. For the "more complicated" issue I think that it is mainly a subjective matter. For example by your answer we need to consider polynomilas, factorization, quadratic equations but the solution can be obtained directly by the definition of trigonometric functions by a simple first order equation. What is simpler? – user Jun 20 '18 at 20:07

5 Answers

By the definition of cosine function we have that

$$\cos \alpha = \cos \theta \iff \alpha = \theta +2k\pi \, \lor \, \alpha = -\theta +2k\pi \quad k\in \mathbb{Z}$$

and thus

$$\cos(3x) = \cos(2x)\iff 3x=2x+2k\pi \, \lor \, 3x=-2x+2k\pi \quad k\in \mathbb{Z}$$

that is

• $x=2k\pi$

• $x=\frac25 k\pi$

• What does that $2k\pi$ mean? How? I still didn't get it. – Hamilton Jun 20 '18 at 15:27
• No, that's now what I meant. – Hamilton Jun 20 '18 at 15:30
• @Hamilton $2k\pi$ is a integer multiple of $2\pi$ – user Jun 20 '18 at 15:30
• Why did you use that method? Can I use it for other trigonometric equations too? – Hamilton Jun 20 '18 at 15:30
• @Hamilton It is a trivial identity from the definition of $\cos x$ indeed by periodicity $\cos x=\cos (x+2k\pi)$ and since $\cos x$ is even $\cos x=\cos (-x+2k\pi)$. You can also see the identity by the trigonometric circle. – user Jun 20 '18 at 15:54

The equality $\cos(3x)=\cos(2x)$ is obviously true when $x=0$ and thus when $t=1.$ Therefore the polynomial $$4t^3-2t^2-3t +1$$ has $t=1$ as one of its zeros. Consequently it can be factored: $$4t^3-2t^2-3t +1 = (t-1)(\cdots\cdots\cdots)$$ The other zeros are those of a quadratic polynomial, written here as $(\cdots\cdots\cdots).$

Hint

You can use sum-product equivalence. Which is:

$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$ so,

$$\cos(3x)-\cos(2x)=0\to-2\sin\left(\frac{3x+2x}{2}\right)\sin\left(\frac{3x-2x}{2}\right)=0$$

$$\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right)=0$$ so,

$$\sin\left(\frac{5x}{2}\right)=0 \text{ or } \sin\left(\frac{x}{2}\right)=0$$

Calling

$$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$

we have

$$e^{3ix}+e^{-3i x} = e^{2ix}+e^{-2i x}$$

or calling $z = e^{ix}$

$$z^6+1 = z^5+z\to z^6-z^5-z+1 = (z-1)^2(z^4+z^3+z^2+z+1) = (z^5-1)(z-1) = 0$$

so the solutions are obvious.

$$x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots$$

To complement gimusi's fine answer, there are other cases when simple methods work:

1. $\cos f(x)=\cos g(x)$
2. $\sin f(x)=\sin g(x)$
3. $\tan f(x)=\tan g(x)$
4. $\sin f(x)=\cos g(x)$
5. $\cot f(x)=\tan g(x)$

where $f(x)$ and $g(x)$ are expressions involving the unknown $x$.

Equation 1 has the solutions $$f(x)=g(x)+2k\pi \qquad\text{or}\qquad f(x)=-g(x)+2k\pi$$

Equation 2 has the solutions $$f(x)=g(x)+2k\pi \qquad\text{or}\qquad f(x)=\pi-g(x)+2k\pi$$

Equation 3 has the solutions $$f(x)=g(x)+k\pi$$ (of course one has also to exclude values of $x$ that make $\tan f(x)$ or $\tan g(x)$ undefined).

Two angles have the same cosine if and only if the points on the unit circle they correspond to have the same $x$-coordinate; two angles have the same sine if and only if the points on the unit circle have the same $y$-coordinate. The $2k\pi$ or $k\pi$ term, with $k$ an integer, represents the periodicity.

What about an equation of the form $\sin f(x)=\cos g(x)$? We can recall that $\sin\alpha=\cos(\pi/2-\alpha)$, so we can reduce it to $$\cos\left(\frac{\pi}{2}-f(x)\right)=\cos g(x)$$ which is type 1 above.

Similarly, $\cot f(x)=\tan g(x)$ can become $$\tan\left(\frac{\pi}{2}-f(x)\right)=\tan g(x)$$ that is, type 3 above.

• Thanks for the quote! Very nice generalization. – user Jun 21 '18 at 9:15