###### Example70

Solve \(\displaystyle \sin(\theta)=\frac{1}{2}\) for all possible values of \(\theta\text{.}\)

To solve this equation, we need to identify all angles, \(\theta\text{,}\) that have a sine value of \(\frac{1}{2}\text{.}\)

In the Inverse Trigonometric Functions Section, we found two solutions to the equation \(\sin(\theta)=\frac{1}{2}\text{,}\)

These solutions are two common angles on the unit circle that have a sine value of \(\frac{1}{2}\text{.}\)

However, if we look at the graph of \(f(\theta)=\sin(\theta)\) to see where it intersects with the horizontal line \(y=\frac{1}{2}\text{,}\) we can see that there are actually more than two solutions to the equation \(\sin(\theta)=\frac{1}{2}\text{.}\)

The two solutions we found above, \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\text{,}\) are shown on the graph below. We call these two angles our initial solutions (or base angles).

Because sine is a periodic function, we can use these two initial solutions to solve for all other solutions. Recall that the period of sine is \(2\pi\text{.}\) Therefore, to find another solution on the graph, we can add \(2\pi\) to \(\frac{\pi}{6}\text{:}\)

Alternatively, we could have also subtracted \(2\pi\) from \(\frac{\pi}{6}\) to give us yet another solution.

These two additional solutions are shown on the graph below.

We could also add or subtract multiples of the period for the other initial solution, \(\frac{5\pi}{6}\text{.}\) For example, if we add or substract one multiple of \(2\pi\) to \(\frac{5\pi}{6}\text{,}\) then the respective solutions are:

We have found all solutions shown on the graph above, but there are still infinitely many solutions left to find because the graph of sine intersects the line \(y=\frac{1}{2}\) infinitely many times.

To find all of the solutions, we add multiples of \(2\pi\) to our initial solutions. That is,

Remember that an integer is a positive or negative whole number, so \(k=..., -2, -1, 0, 1, 2, ...\)

We can relate these solution sets back to the intersection points on the graph above. When \(k=0\text{,}\) we get our two initial solutions:

and

When \(k=1\text{,}\) we get

and

This corresponds to the two solutions to the right of our initial solutions on the graph.

When \(k=-1\text{,}\) we get the two solutions to the left of our initial solutions: \(-\frac{11\pi}{6}\) and \(-\frac{7\pi}{6}\text{,}\) and so on.

We can also relate these solution sets to angles on the unit circle.