Find all solutions to the following equation:

$$4\cos^2 \theta - 8 \cos \theta + 3 = 0$$

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

Here, this involves solving a quadratic equation (i.e., $4x^2 - 8x + 3 = 0$). Indeed, because of this, we say that our given equation is "quadratic in terms of $\cos \theta$".

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

$$\begin{array}{rcll} 4\cos^2 \theta - 8 \cos \theta + 3 &=& 0 & \scriptsize{\textrm{seeing zero on one side, we attempt to factor the other side}}\\\\ (2 \cos \theta - 1)(2 \cos \theta - 3) &=& 0 & \overset{\normalsize{\textrm{one of these factors must then be zero,}}}{\scriptsize{\textrm{which gives us two simpler equations to solve...}}} \end{array}$$

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

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

So the only possible values for the cosine of the angle(s) we seek are $1/2$ and $3/2$.

One of these we can reject out of hand. To see this, recall that the range of the cosine function is $[-1,1]$. As $3/2$ is outside of this interval, there is no angle $\theta$ that satisfies $\cos \theta = 3/2$. So our second "linear" equation is never satisfied, and contributes no solution.

However, recalling that a cosine is an $x$-coordinate of a point on the unit circle, the other cosine value tells us that we seek an angle associated with a point on the unit circle whose $x$-coordinate is $1/2$.

Within one full counter-clockwise rotation (i.e, $0 \le \theta \lt 2\pi$), there are two such points as the below drawing suggests.

Accounting for angles co-terminal to these (i.e., those whose measures differ from these two by $2\pi n$ for integers $n$), we conclude that all solutions to $4\cos^2 \theta - 8 \cos \theta + 3 = 0$ are given by

$$\theta = \frac{\pi}{3} \pm 2\pi n \,\, , \,\, \frac{5\pi}{3} \pm 2\pi n \quad \textrm{where } n \textrm{ is an integer}$$

We could stop there, although we can take advantage of symmetry to write our solution set in a more succinct way. Note that every angle co-terminal to $5\pi/3$ must also be co-terminal to $-\pi/3$ and vice-versa.


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