B. $30^\circ, 60^\circ$ Step 1: Find the principal values for $2x$. Given the equation $\sin 2x = \frac{\sqrt{3}}{2}$. The reference angle for $\sin \theta = \frac{\sqrt{3}}{2}$ is $60^\circ$. Since $\sin 2x$ is positive, $2x$ must be in Quadrant I or Quadrant II. The principal values for $2x$ are: $$2x = 60^\circ$$ or $$2x = 180^\circ - 60^\circ = 120^\circ$$ Step 2: Find the general solutions for $2x$. The general solutions for $\sin \theta = k$ are $\theta = \alpha + 360^\circ n$ or $\theta = (180^\circ - \alpha) + 360^\circ n$, where $\alpha$ is the reference angle and $n$ is an integer. So, for $2x$: $$2x = 60^\circ + 360^\circ n$$ or $$2x = 120^\circ + 360^\circ n$$ Step 3: Solve for $x$. Divide both general solutions by 2: $$x = \frac{60^\circ}{2} + \frac{360^\circ n}{2} \implies x = 30^\circ + 180^\circ n$$ or $$x = \frac{120^\circ}{2} + \frac{360^\circ n}{2} \implies x = 60^\circ + 180^\circ n$$ Step 4: Apply the given range $0^\circ \le x \le 180^\circ$. For $x = 30^\circ + 180^\circ n$: • If $n=0$, $x = 30^\circ + 180^\circ(0) = 30^\circ$. This is within the range. • If $n=1$, $x = 30^\circ + 180^\circ(1) = 210^\circ$. This is outside the range. For $x = 60^\circ + 180^\circ n$: • If $n=0$, $x = 60^\circ + 180^\circ(0) = 60^\circ$. This is within the range. • If $n=1$, $x = 60^\circ + 180^\circ(1) = 240^\circ$. This is outside the range. The solutions for $x$ in the given range are $30^\circ$ and $60^\circ$. The final answer is $\boxed{\text{B. } 30^\circ, 60^\circ}$.