When is $\cos (x) \geq \frac{1}{2}$? When is $\cos (x) \geq \frac{1}{2}$?
I know the function repeats, so I know I should end up with an interval that allows for integer multiples.
e.g. something like this (but obviously not this exactly) $[0 + n, \pi + n]$.
 A: First, we solve $\cos x=\frac12$, which is for $x=\frac{\pi}{3}$ and $x=\frac{5\pi}{3}$.
Then, we take a look at the unit circle.

The cosine value is the $x$-coordinate, so the question we ask is: "for which $x$ is the $x$-coordinate greater than or equal to $\frac12$?".
In the first quadrant, we have $0\leq x\leq\frac{\pi}{3}$ or $[0,\frac{\pi}{3}]$.
In the fourth quadrant, we have $\frac{5\pi}{3}\leq x\leq2\pi$ or $[\frac{5\pi}{3},2\pi]$.
Now, we add the periodicity. We now that $\cos(x)=\cos(x+2\pi)$, so we can always add $+2\pi$ to our answer.
Therefore, the final answer is:
$\cos(x)\geq\frac12$ for $x\in[2\pi n,\frac{\pi}{3}+2\pi n]\cup[\frac{5\pi}{3}+2\pi n,2\pi+2\pi n]$ or, preferably for some, $x\in[-\frac{\pi}{3}+2\pi n, \frac{\pi}{3}+2\pi n]$, ($n\in\mathbb{Z}$)
A: $[-\frac{\pi}{3}+2\pi n, \frac{\pi}{3} + 2\pi n]$
On $[-\pi, \pi], \cos(x) > 1/2 \quad \iff \quad x\in[-\pi/3, \pi/3]$
Indeed, $\cos(\pi/3) = 1/2$, and you know that cosine is an even function.
Moreover, cosine is $2\pi$-periodic, and therefore have you can translate $[-\pi/3, \pi/3]$ by $2\pi n, n\in \mathbb{N}$ to get all possible answers.
A: Some hints:


*

*For what values of $x$ is the $x$ component of a point on the unit circle at least $1/2$?  (Drawing this may help.)

*You know the function repeats.  How much do you add to the angle to go once around the circle?

*How do you express this amount so that you can go around twice, three times, $n$ times?  How about going around in the other direction?

A: Counting the angles in turns ($360°$), and looking at the plot of the $\cos$ function ($\cos(\pm60°)=\frac12$), the intervals are $$[-\frac16+n,\frac16+n].$$

