# How to solve for angles $4\theta = \theta$?

I want to find all the angles in $[0, 2\pi)$ for which $4\theta = \theta$ is true. I can obviously get $\theta = 0$, but the other solutions are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. How do I find these particular ones?

• @graydad: He never claimed that those two angles are equal, but that each of them individually satisfies $4\theta\equiv\theta \pmod{2\pi}$. Commented Mar 23, 2015 at 20:25
• @HenningMakholm The $\pmod{2\pi}$ part got left out of the question. I wondered if the problem was it also got left out of the thinking about the question, or whether the problem was dealing with the modulus. Commented Mar 23, 2015 at 20:28
• @DavidK: My immediate interpretation of the question was that the OP understood that $\frac{2\pi}3$ and $\frac{4\pi}3$ are solutions, but couldn't figure out how he would have arrived at them systematically. Commented Mar 23, 2015 at 20:30
• @HenningMakholm That too was certainly possible based on the problem statement. Commented Mar 23, 2015 at 20:33

Two angles are "the same angle" when they differ by a multiple of $2\pi$. So you want to solve $$4\theta = \theta+2k\pi$$ for all $k$. Solving this, we get $$\theta = \frac k3 2\pi$$ from which we see that solutions for $k$s that differ by a multiple of $3$ will be the same. So we get all solutions by taking the ones for $k=0,1,2$.

You can set up the equations $$4\theta=\theta+2\pi$$ and $$4\theta=\theta+4\pi$$ to get the two other solutions you are after. You could add $6\pi$ but the solution to that would fall outside your bounds.

The writing $4\theta = \theta$ is not accurate. I assume that you are solving some equation that takes the form: $\cos(4\theta) = \cos(\theta)$ or $\sin(4\theta) = \sin(\theta)$ or possibly something else.

This boils down to finding the solution of $4\theta \equiv \theta \pmod {2\pi}$

$$4\theta \equiv \theta \pmod {2\pi} \iff \exists \ k \in \mathbb Z \ / \ 3\theta = 2k\pi \iff \exists \ k \in \mathbb Z \ / \ \theta = \frac23k\pi$$

Thus, the solutions to this equation are the elements of $\{\frac23k\pi \ ; \ k \in \mathbb Z\}$.

Then, look for those solutions which are in $[0,2\pi)$.