Originally Posted by

**Grandad** Hello thereddevils

As red_dog has said, you need to solve two equations

$\displaystyle \sin2x = \frac12$

and

$\displaystyle \sin2x=-\frac12$

Since you want values of $\displaystyle x$ between $\displaystyle -\pi$ and $\displaystyle \pi$, you'll need to look at values of $\displaystyle 2x$ between $\displaystyle -2\pi$ and $\displaystyle 2\pi$.

I always find the graph of $\displaystyle \sin\theta$ easier to use than the unit circle, so look at the sketch I've attached. The easiest value is, of course, $\displaystyle \sin(\tfrac{\pi}{6})=\tfrac12$. Look at where the dotted lines intersect the graph. They are at:

$\displaystyle -\frac{11\pi}{6},-\frac{7\pi}{6},-\frac{5\pi}{6},-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7 \pi}{6},\frac{11\pi}{6}$

So these are the possible values of $\displaystyle 2x$. Divide them all by $\displaystyle 2$, and you're done.

Grandad