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!
 A: 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 $(\cdots\cdots\cdots).$
A: 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,
$$\sin\left(\frac{5x}{2}\right)=0 \text{ or } \sin\left(\frac{x}{2}\right)=0$$
A: 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.
$$
x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots
$$
A: To complement gimusi's fine answer, there are other cases when simple methods work:


*

*$\cos f(x)=\cos g(x)$

*$\sin f(x)=\sin g(x)$

*$\tan f(x)=\tan g(x)$

*$\sin f(x)=\cos g(x)$

*$\cot f(x)=\tan g(x)$


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.
A: 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


*

*$x=2k\pi$

*$x=\frac25 k\pi$
