1
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ @graydad: He never claimed that those two angles are equal, but that each of them individually satisfies $4\theta\equiv\theta \pmod{2\pi}$. $\endgroup$ Commented Mar 23, 2015 at 20:25
  • 1
    $\begingroup$ @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. $\endgroup$
    – David K
    Commented Mar 23, 2015 at 20:28
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Mar 23, 2015 at 20:30
  • $\begingroup$ @HenningMakholm That too was certainly possible based on the problem statement. $\endgroup$
    – David K
    Commented Mar 23, 2015 at 20:33

3 Answers 3

5
$\begingroup$

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 $k=0,1,2$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $[0,2\pi)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .