Given the equation 2x = sqrt(3)2.

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

$$2x = 180^\circ - 60^\circ = 120^\circ$$

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

$$2x = 120^\circ + 360^\circ n$$

Divide both general solutions by 2:

$$x = \frac{60^\circ}{2} + \frac{360^\circ n}{2} \implies x = 30^\circ + 180^\circ n$$

or

$$x = \frac{120^\circ}{2} + \frac{360^\circ n}{2} \implies x = 60^\circ + 180^\circ n$$

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 $\text{B. } 30^\circ, 60^\circ$.