# Solution

Find all solutions to the following equation

$$2\sin^2 \theta + (2- \sqrt{3}) \sin \theta - \sqrt{3} = 0$$

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 solving a quadratic equation [i.e., $2x^2 + (2-\sqrt{3})x - \sqrt{3} = 0$]. Equivalently, our given equation is "quadratic in terms of $\sin \theta$".

We solve this "quadratic" equation in the usual way:

$$\begin{array}{rcll} 2\sin^2 \theta + (2-\sqrt{3})\sin \theta - \sqrt{3} &=& 0 & \scriptsize{\textrm{seeing zero on one side, we attempt to factor the other side}}\\\\ (2\sin \theta - \sqrt{3})(\sin \theta + 1) &=& 0 & \overset{\normalsize{\textrm{one of these factors must then be zero,}}}{\scriptsize{\textrm{which gives us two simpler equations to solve...}}} \end{array}$$

As seen below, these two simpler equations are both "linear" in terms of $\sin \theta$, so we solve them accordingly.

$$\begin{array}{rcl|rcl} 2\sin \theta - \sqrt{3} &=& 0 \quad & \quad \sin \theta + 1 &=& 0\\\\ 2\sin \theta &=& \sqrt{3} \quad & \quad \sin \theta &=& -1\\\\ \sin \theta &=& \frac{\sqrt{3}}{2} & & & \end{array}$$

So the only possible values for the sine of the angle(s) we seek are $\sqrt{3}/2$ and $-1$.

Recall, a sine value is a $y$-coordinate -- a height -- of a point on the unit circle associated with a given angle. As such, we seek points on the unit circle at height $\sqrt{3}/2$ and $-1$.

Within one full counter-clockwise revolution (i.e, where $0 \le \theta \lt 2\pi$), there are three such points, where $\theta = \pi/3$, and $2\pi/3$, and $3\pi/2$, as shown in the picture below:

Accounting for angles co-terminal to these three (i.e., those whose measures differ from these by $2\pi n$ for integers $n$), we conclude that all solutions to $2\sin^2 \theta + (2- \sqrt{3}) \sin \theta - \sqrt{3} = 0$ are given by

$$\theta = \frac{\pi}{3} \pm 2\pi n \, \, , \, \, \frac{2\pi}{3} \pm 2\pi n \, \, , \, \, or \, \, \frac{3\pi}{2} \pm 2\pi n \quad \textrm{where } n \textrm{ is an integer}$$