MathematicsPreCalculus Mathematics at Nebraska

We solved this equation for all possible values of \(\theta\) in the previous section. We found that

To find values of \(\theta\) in the interval \([-\pi,\pi]\text{,}\) we must test values of \(k\text{.}\) To better organize our work, we will create a "\(k\)-table" where we can keep track of different values of \(k\text{,}\) the angle \(\theta\) they yield, and whether or not that angle is in the interval \([-\pi,\pi]\text{.}\)

Notice that for both positive and negative values of \(k\text{,}\) we stopped as soon as we reached a value of \(\theta\) that was outside of the interval \([-\pi,\pi]\text{.}\) This is because for positive values of \(k\text{,}\) \(\theta\) was getting larger, so once \(\theta\) was greater than \(\pi\) it would always be greater than \(\pi\text{.}\) Similarly, when \(k\) was negative, \(\theta\) was getting more negative, so once it was less than \(-\pi\) it would always be less than \(-\pi\text{.}\) Therefore, from the table we can see that there are exactly six solutions in the interval \([-\pi,\pi]\text{:}\)

Let's start by noting that \(\sin(\theta)\cos(\theta) - \frac{1}{2}\cos(\theta)\) has a greatest common factor of \(\cos(\theta)\text{.}\) Then we can we can rewrite the equation as

This means that \(\cos(\theta) = 0\) or \(\sin(\theta) - \frac{1}{2} = 0\text{.}\)

If \(\cos(\theta) = 0\text{,}\) then from the unit circle we find that \(\theta = \frac{\pi}{2}+\pi k\) where \(k\) is any integer.

If \(\sin(\theta) - \frac{1}{2} = 0\text{,}\) then we are solving \(\sin(\theta) = \frac{1}{2}\text{.}\) From the unit circle, two initial solutions are given by \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\text{.}\) Since \(\sin(\theta)\) has period \(2\pi\text{,}\) we have then that the solutions are given by

and

Therefore, the solutions to the original equation are given by

where \(k\) is any integer.

Consider the function \(f(x)=3x^2-5x\text{.}\) If you were asked to find all solutions to \(f(x)=2\text{,}\) you could use algebra to arrive at two solutions:

Therefore, we get the two solutions

Following these steps as above with \(\sec(\theta)\) instead of \(x\text{,}\) we know that either

or equivalently, either

There are no solutions to the equation \(\cos(\theta)=-3\text{,}\) so we need only consider \(\cos(\theta)=\frac{1}{2}\text{.}\) Solving the equation \(\cos(\theta)=\frac{1}{2}\text{,}\) we get two solution sets. First,

Using symmetry, the other solution set for \(\cos(\theta)=\frac{1}{2}\) has base angle \(-\frac{\pi}{3}\text{,}\) which gives

Finally, we need to determine which solutions lie in the interval \(0\le\theta\le 2\pi\text{.}\) We will do this by constructing a table of \(k\)-values, starting at \(k=0\) and increasing \(k\) until we are outside of the interval, and then at \(k=-1\) and decreasing until we are outside of the interval.

Consider the function \(f(x)=2x^2+3x\text{.}\) If you were asked to find all solutions to \(f(x)=-1\text{,}\) you could use algebra to arrive at two solutions:

Therefore, we get the two solutions

Following these steps as above with \(\sin(\theta)\) instead of \(x\text{,}\) we know that either

We can use the unit circle to solve \(\sin(\theta)=-\frac{1}{2}\text{.}\) There are two initial solutions, namely \(\theta=\frac{7\pi}{6}\) and \(\theta=\frac{11\pi}{6}\text{.}\) This leads to the following two solution sets:

We can use the unit circle to solve \(\sin(\theta)=-1\) as well. There is just one initial solution, \(\theta=\frac{3\pi}{2}\text{.}\) Thus we get a third solution set

We now need to determine which solutions lie in the interval \(0\le\theta\le 2\pi\text{.}\) We will do this by constructing a table of \(k\)-values, starting at \(k=0\) and increasing \(k\) until we are outside of the interval, and then at \(k=-1\) and decreasing until we are outside of the interval.

Therefore there are exactly three solutions to the original equation \(2\sin^2(\theta)+3\sin(\theta)=-1\) in the interval \([0,2\pi]\text{:}\)