Find all solutions to the following equation

$$2 \sin \theta = 1$$

The only variable occurs in a single trigonometric function, $\sin \theta$. Thus, we attempt to solve for this $\sin \theta$. That is to say, we treat this $\sin \theta$ itself as some unknown variable $x$ and try then to solve for that $x$.

Here, this involves only solving a simple, linear equation (i.e., $2x = 1$). Indeed, because of this, we say that our given equation is "linear in terms of $\sin \theta$".

We solve this "linear" equation in the usual way -- in this case, by dividing by $2$,

$$\begin{array}{rcc} 2\sin \theta &=& 1\\ \sin \theta &=& \frac{1}{2} \end{array}$$

From here -- if we are lucky -- we will be able to determine which values of $\theta$ satisfy the above using the unit circle.

Remember, the value of the sine function is a $y$-coordinate -- a "height" -- of some point on the unit circle. As such, we look for points at height $1/2$ on the unit circle.

Examining one full counter-clockwise revolution ($0 \le \theta \lt 2\pi$), there are two such points which correspond to angles $\theta = \pi/6$ and $\theta = 5\pi/6$.

Note, we can't forget all of the angles co-terminal to these two angles, as they share the same sine values as the two angles above and would thus also be solutions to our equation.

To account for these, we recall two angles are co-terminal if they differ by some number of full revolutions. That is to say, co-terminal angles differ by $2\pi n$ radians, where $n$ represents the integer number of full revolutions involved.

Thus, all solutions to $2\sin x = 1$ are given by:

$$\theta = \frac{\pi}{6} \pm 2\pi n \,\, , \,\, \frac{5\pi}{6} \pm 2\pi n \quad \textrm{where } n \textrm{ is an integer}$$