How do you find two solutions to $\sin (x) = \dfrac{1}{2}$ ?

Hint: sin(x) is positive in the first and second quadrant. The range in which the solution can be found is $[0,\pi ]$. Using this information and the result $\sin (x) = \sin (\pi - x)$, the solutions for $\sin (x) = \dfrac{1}{2}$ can be found.

Complete Step by Step Solution:
The given expression in the question is $\sin (x) = \dfrac{1}{2}$. We need to find the solutions in the interval $[0,2\pi ]$. This implies that we need to look for angles in this particular interval such that the sine of these angles would be $\dfrac{1}{2}$.
We know that the angle satisfying the expression cannot be in the third or the fourth quadrant because the sine function is negative in these quadrants. So, we can reduce the range of finding the solution from $[0,2\pi ]$ to $[0,\pi ]$.
A known result is $\sin (30^\circ ) = \dfrac{1}{2}$. Thus, one of the solutions of x is $30^\circ $.
To find the other angle that would satisfy the expression, we can use the result $\sin (x) = \sin (\pi - x)$
So, according to the symmetry of the sine graph, we can find the value of the second angle by substituting x as $30^\circ $.
Therefore, $\sin (30^\circ ) = \sin (180^\circ - 30^\circ ) = \sin (150^\circ )$
As we know that $\sin (30^\circ ) = \dfrac{1}{2}$ , we can deduce that $\sin (150^\circ ) = \dfrac{1}{2}$.

The solutions of $\sin (x) = \dfrac{1}{2}$ are $x = 30^\circ ,150^\circ $.

Note:
We can prove the result $\sin (x) = \sin (\pi - x)$ in the following manner:
$\sin (\pi - x)$ can be expanded using the sine formula: $\sin (a - b) = \sin (a)\cos (b) - \cos (a)\sin (b)$
Hence, $\sin (\pi - x) = \sin (\pi )\cos (x) - \cos (\pi )\sin (x)$
Substituting $\sin (\pi ) = 0$ and $\cos (\pi ) = - 1$ , we get $\sin (\pi - x) = (0)\cos (x) - ( - 1)\sin (x) = \sin (x)$
$\therefore \sin (\pi - x) = \sin (x)$