Answer
The solution set is $\{\varnothing\}$
Work Step by Step
$$\sin^2\frac{x}{2}-2=0$$ over interval $[0,2\pi)$
1) Find corresponding interval for $\frac{x}{2}$
As the interval for $x$ is $[0,2\pi)$, we can only write it as the inequality
$$0\le x\lt2\pi$$
Therefore, $$0\le\frac{x}{2}\lt\pi$$
So the interval for $\frac{x}{2}$ is $[0,\pi)$.
2) Now consider back the equation $$\sin^2\frac{x}{2}-2=0$$
$$\sin^2\frac{x}{2}=2$$
$$\sin\frac{x}{2}=\pm\sqrt2$$
Recall the defined range of $\sin\theta$ is $[-1,1]$.
Here we need to find values of $\frac{x}{2}$ so that $\sin\frac{x}{2}=\pm\sqrt2$. However, as $\sqrt2\gt1$ and $-\sqrt2\lt(-1)$, it follows that $\pm\sqrt2\notin[-1,1]$, the range of sine. That means there is no value of $\frac{x}{2}$ that $\sin\frac{x}{2}$ could equal $\pm\sqrt2$
That leads to no values of $x$ which could satisfy the given equation.
In other words, the solution set is $\{\varnothing\}$
