Answer
The equation has the solution set: $$\{\frac{\pi}{2},\frac{3\pi}{2}\}$$
Work Step by Step
$$\sin\frac{x}{2}=\sqrt2-\sin\frac{x}{2}$$ over interval $[0,2\pi)$
1) Find corresponding interval for $\frac{x}{2}$
Interval $[0,2\pi)$ can be written as
$$0\le x\lt2\pi$$
That means, for $\frac{x}{2}$, the interval would be
$$0\le\frac{x}{2}\lt\pi$$
or $$\frac{x}{2}\in[0,\pi)$$
2) Now consider back the equation $$\sin\frac{x}{2}=\sqrt2-\sin\frac{x}{2}$$
$$2\sin\frac{x}{2}=\sqrt2$$
$$\sin\frac{x}{2}=\frac{\sqrt2}{2}$$
Over the interval $[0,\pi)$, there are 2 values whose $\sin$ equals $\frac{\sqrt2}{2}$, which are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, meaning that
$$\frac{x}{2}=\{\frac{\pi}{4},\frac{3\pi}{4}\}$$
So $$x=\{\frac{\pi}{2},\frac{3\pi}{2}\}$$
