Efficiently finding $\theta$ such that $\cos\theta = -\frac12$ and $\sin\theta = \frac{\sqrt{3}}{2}$. I know the quadrant, but what are the angles? I'm having trouble solving trigonometric equations. For example, let's say I'm solving a problem and I arrive at a trigonometric equation that says, 
$$\cos\theta = -\frac12 \quad\text{and}\quad \sin\theta = \frac{\sqrt{3}}{2} $$ 
At this point, I get stuck and I don't have an efficient way to proceed apart from picking up a calculator. 
I can figure that the quadrants (from the signs of the ratios) -- but I can't figure out the angles. What is a good way to figure out the angle? Specifically, how do I systematically solve $\sin$, $\cos$, and $\tan$ trigonometric equations? (I can reciprocate the other three into these ratios.)
I don't have trouble figuring out angles between $0^\circ$ to $90^\circ$ (since I have that memorized), but for angles in other quadrants, I get stuck.
 A: If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)
Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)
You can use the following identities (which are derived from the aforementioned facts).

$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$
$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$
$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$
$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$
$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$
$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$
$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$
$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$
$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$
$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$

I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.
For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$
You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.
$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$
$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$
This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.
A: Consider the following equilateral triangle $ABC$, where $D$ is the foot of the altitude $\overline{CD}$

Let $[AB]=a$
Thus $$cos\bigl(\angle DAC \bigr)=cos\Bigl(\frac{\pi}{3}\Bigr)=\frac{AC}{AD}=\frac{1}{2}$$
Consider now the trigonometric identities $$cos\bigl(\pi-\theta\bigr)=-cos\bigl(\theta\bigr)\Rightarrow cos\Bigl(\frac{2\pi}{3}\Bigr)=-\frac{1}{2}$$
$$cos\bigl(\theta+2\pi\bigr)=cos(\theta) \Rightarrow cos\Bigl(\frac{2\pi}{3}+2\pi n\Bigr)=-\frac{1}{2}\; \forall n\in \mathbb{Z}$$
Now back to the equilateral triangle, observe that by the Pythagorean theorem
$$[CD]=\sqrt{{[AC]^2}-{[AD]^2}}=\sqrt{{a^2}-{\Bigl(\frac{a}{2}\Bigr)^2}}=\frac{\sqrt{3}}{2}*a$$ Hence $$sin\bigl(\angle DAC\bigr)=sin\Bigl(\frac{\pi}{3}\Bigr)=\frac{[DC]}{[AC]}=\frac{\sqrt{3}}{2}$$
Since $sin(\theta)=sin(\theta + 2\pi)$
$$sin\Bigl(\frac{\pi}{3}+2n\pi\Bigr)=\frac{\sqrt{3}}{2} \; \forall n\in \mathbb{Z}$$
$\mathbf{PS}$
Keeping in mind the sine- (red) and cosine (green) functions might help with this problems
 
A: 
In the above, $\theta$ is measured anticlockwise from the positive horizontal axis and P is the angle with the nearest horizontal axis. P takes values from 0 to $\frac{\pi}{2}$. 
Suppose you want to solve $cos(\theta) = -0.5$,


*

*Solve $cos(P) = +0.5$ using inverse cosine on your calculator. $P =\frac{\pi}{3}$  is called the reference angle. 

*Ask yourself where $cos(\theta)$ is negative. This is in the second and third quadrants. In the second quadrant $\theta = \pi-P = \frac{2\pi}{3}$ and in the third quadrant $\theta = \pi+P = \frac{4\pi}{3}$.

*If you want more solutions then solve $\theta = \pi-P+2k\pi$ and $\theta = \pi+P+2k\pi$ instead, where k is any positive or negative integer.


eg2. Solve $\sin(3\theta+1) = 0.5$ for $0\le \theta \le 2\pi$


*

*Solve $\sin(P) = 0.5$ then $P = \frac{\pi}{3}$

*Where is $sin$ positive? In quadrant 1 so put $3\theta+1 = P+2k\pi$ and in quadrant 2 so put $3\theta+1 = \pi-P+2k\pi$.
Now solve for $\theta$ and keep any solutions which fall in the range $0\le \theta \le 2\pi$.
