###### Example70

Solve the trigonometric equation \(2\sin(3\theta)=1\) for all possible values of \(\theta\text{.}\)

Similar to the examples in Section, we can sketch a graph of \(f(\theta)=2\sin(3\theta)\) to see where it intersects with the line \(y=1\text{.}\) Just like before, the angles that correspond to the intersection points of these two functions are the solutions to the equation \(2\sin(3\theta)=1\text{.}\) If we look at the graph, we can see that there are an infinite number of solutions to this equation.

The key difference between this example and the examples in Section is that we now have an input of \(3\theta\) instead of just \(\theta\) into our sine function. To solve this equation, we need to identify all angles, \(\theta\text{,}\) such that \(2\sin(3\theta)=1\text{.}\)

To solve this problem, we can first isolate the sine function by dividing each side of the equation by 2, which gives us

While we know what angles have a sine value of \(1/2\text{,}\) it is less clear how to proceed when we have \(3\theta\) as our input. To make the problem more familiar, we can introduce a new variable \(\phi\) and set \(\phi=3\theta\text{.}\) Then, substituting \(\phi\) into our equation, we get

In Example63, we found that the solutions to this equation are

Since we want to solve for all possible values of \(\theta\text{,}\) we can use our substitution and replace \(\phi\) with \(3\theta\) in the solutions above. Therefore, we have that

Our final step in solving for \(\theta\) is to divide each side of the equations by 3, so we have

and simplifying we get

Notice that when we divide the right side of our equations by 3, we must divide each term by 3 in order to simplify the equations. Therefore, we end up adding multiples of \(2\pi/3\) to our initial solutions instead of multiples of \(2\pi\text{.}\) Also note that the sine function we are dealing with, \(f(\theta)=2\sin(3\theta)\text{,}\) has a period of \(2\pi/3\text{.}\)