I am just trying to figure out what formula I would use to solve this equation. The problem is solve $\cos(3\theta)=1/2$; for all $0\leq \theta\leq 360^\circ$. I want to say I would use the double angle formula but I am not positive.
[Math] Double Angle Formula
trigonometry
Related Solutions
I'm not really sure why you need to use double angle formula (some exercise?), but I'd just do it in most straightforward way \begin{align} 3 \tan \theta &= 2 \cos \theta \\ 3 \sin \theta &= 2 \cos^2 \theta = 2 - 2\sin^2 \theta \end{align} $$ 2\sin^2 \theta + 3 \sin \theta - 2 = 0 \\ \sin \theta = \frac {-3 \pm \sqrt{9+16}}{4} = \frac {-3 \pm 5}{4} = \left [ \begin{array}{l} -2 \text{ (spurious solution)}\\ \frac 12 \text{ (correct solution)} \end{array}\right . $$ So, final answer is $\theta \in \left \{ \frac \pi 6, \frac {5 \pi}6\right \}$ or $\theta \in \left \{ 30^\circ, 150^\circ\right \}$ if operating with degrees.
Omission of the negative sign is not necessary.
The technique you described in your question involves finding a reference angle in the first quadrant (the reason you drop the negative sign), determining the quadrants in which the cosine is negative, and then finding the angles in those quadrants with that reference angle.
In this case, the reference angle is $60^\circ$, the quadrants in which the cosine is negative are the second and third quadrants (since the $x$-coordinate is negative), and the angles in those quadrants with a reference angle of $60^\circ$ are, respectively, $180^\circ - 60^\circ = 120^\circ$ and $180^\circ + 60^\circ = 240^\circ$.
While this method works, it is not necessary.
The arccosine function (inverse cosine function) is defined by $$\arccos x = \theta \iff x = \cos\theta, 0 \leq \theta \leq \pi$$ where $\theta$ is measured in radians. Since $2\pi/3$ is the unique angle $\theta$ in the interval $[0, \pi]$ such that $\cos\theta = -1/2$, $$\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$$ Thus, one solution of the equation $\cos\theta = -1/2$ is $2\pi/3$, which is $120^\circ$. Using symmetry, we obtain $$\cos\theta = \cos(2\pi - \theta)$$ since the $x$-coordinates of the points where the terminal sides of the angles $\theta$ and $2\pi - \theta$ intersect the unit circle are equal. (The angle $\theta$ in the diagram is not the same as the one in the problem.)
Hence, another solution is $$2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$$ which is $240^\circ$.
Best Answer
Use the fact that $\cos{x}=\cos{y}\iff x=\pm y+2\pi n$.
So we have $$\cos{3\theta}=\frac{1}{2}=\cos\left(\frac{\pi}{3}\right)\\\implies3\theta=\pm\frac{\pi}{3}+2\pi n\\\implies\theta=\pm\frac{\pi}{9}+\frac{2\pi n}{3}$$