At the angle $\frac\pi3$ on the unit circle, the $x$ coordinate
is not merely "about $\frac12$"; it is exactly $\frac12$.
So you have already found one solution, $\theta = \frac\pi6$.
If you're looking for points where $x$ is $\frac12$, a good place to look
is the set of all points that satisfy $x=\frac12$. That set of points is
a vertical line that crosses the $x$ axis at the coordinate $\frac12$.
If this is belaboring obvious facts, I apologize; but the next step is
to ask, how many times does that line intersect the unit circle?
You found one point of intersection already, and measured the angle
to it and found it was $\frac\pi3$; are there any others?
(By the way, this works for solving problems like $\sin(\alpha) = \frac12$,
too. If you want to find points with sine $\frac12$, you're looking for
points on the horizontal line $y=\frac12$.)
Another fact that you probably know, but is worth repeating:
whenever you add or subtract $2\pi$ to or from an angle, you go exactly
once around the unit circle and end up exactly where you were.
So if $\cos(2\cdot\frac\pi6) = \frac12$ then also
$\cos(2\cdot\frac\pi6 + 2\pi) = \frac12$,
because the angles $2\cdot\frac\pi6$ and $2\cdot\frac\pi6 + 2\pi$ point at exactly the same point. So do $2\cdot\frac\pi6 - 2\pi$, $2\cdot\frac\pi6 + 4\pi$, $2\cdot\frac\pi6 - 4\pi$, $2\cdot\frac\pi6 + 6\pi$, and so forth.
And if each of those numbers in the preceding list is a possible value
of $2\theta$, what possible values of $\theta$ do they represent?
Best Answer
HINT
Are you aware of the identity \begin{align*} \cos(2x) = 1 - 2\sin^{2}(x) \end{align*}
You can apply it in order to obtain a quadratic equation.
Another possible approach consists in noticing that \begin{align*} -\sin(x) = \sin(-x) = \cos\left(\frac{\pi}{2} + x\right) \end{align*}
Can you take it from here?