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

Question

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!

Answer 1

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

• $x=2k\pi$

• $x=\frac25 k\pi$

Answer 2

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.

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

Answer 3

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,

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

Answer 4

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 $(\cdots\cdots\cdots).$

Answer 5

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.