# How do you find all solutions trigonometric equations?

Mar 9, 2015

As a general description, there are 3 steps. These steps may be very challenging, or even impossible, depending on the equation.

Step 1: Find the trigonometric values need to be to solve the equation.
Step 2: Find all 'angles' that give us these values from step 1.
Step 3: Find the values of the unknown that will result in angles that we got in step 2.

(Long) Example
Solve: $2 \sin \left(4 x - \frac{\pi}{3}\right) = 1$

Step 1: The only trig function in this equation is $\sin$.
Sometimes it is helpful to make things look simpler by replacing, like this:
Replace $\sin \left(4 x - \frac{\pi}{3}\right)$ by the single letter $S$. Now we need to find $S$ to make $2 S = 1$. Simple! Make $S = \frac{1}{2}$
So a solution will need to make $\sin \left(4 x - \frac{\pi}{3}\right) = \frac{1}{2}$

Step 2: The 'angle' in this equation is $\left(4 x - \frac{\pi}{3}\right)$. For the moment, let's call that $\theta$. We need $\sin \theta = \frac{1}{2}$
There are infinitely many such $\theta$, we need to find them all.

Every $\theta$ that makes $\sin \theta = \frac{1}{2}$ is coterminal with either $\frac{\pi}{6}$ or with $\frac{5 \pi}{6}$. (Go through one period of the graph, or once around the unit circle.)
So $\theta$ Which, remember is our short way of writing $4 x - \frac{\pi}{3}$ must be of the form: $\theta = \frac{\pi}{6} + 2 \pi k$ for some integer $k$ or of the form $\theta = \frac{5 \pi}{6} + 2 \pi k$ for some integer $k$.

Step 3:
Replacing $\theta$ in the last bit of step 2, we see that we need one of: $4 x - \frac{\pi}{3} = \frac{\pi}{6} + 2 \pi k$ for integer $k$
or $4 x - \frac{\pi}{3} = \frac{5 \pi}{6} + 2 \pi k$ for integer $k$.

Adding $\frac{\pi}{3}$ in the form $\frac{2 \pi}{6}$ to both sides of these equations gives us:
$4 x = \frac{3 \pi}{6} + 2 \pi k = \frac{\pi}{2} + 2 \pi k$ for integer $k$ or
$4 x = \frac{7 \pi}{6} + 2 \pi k$ for integer $k$.

Dividing by $4$ (multiplying by $\frac{1}{4}$) gets us to:

$x = \frac{\pi}{8} + \frac{2 \pi k}{4}$ or
$x = \frac{7 \pi}{24} + \frac{2 \pi k}{4}$ for integer $k$.

We can write this in simpler form:
$x = \frac{\pi}{8} + \frac{\pi}{2} k$ or
$x = \frac{7 \pi}{24} + \frac{\pi}{2} k$ for integer $k$.

Final note The Integer $k$ could be a positive or negative whole number or 0. If $k$ is negative, we're actually subtracting from the basic solution.