Exercise 61 Page 917

Recall that the length of the radius of a unit circle is $1.$ Also, the $x\text{-}$coordinate of the point $P$ is $\cos \theta$ and the $y\text{-}$coordinate of $P$ is $\sin \theta .$ Based on the unit circle, we can determine the values of $\cos \theta$ and $\sin \theta$ where $\theta$ is equal to $\pi .$ We will now examine the unit circle again to see if the equations $\cos \theta =\text{-}1$ and $\sin \theta =0$ have any solutions other than $\pi .$

Solution(s) of $\sin \theta =0$

Let's consider $\sin \theta =0.$ Remember, the $y\text{-}$coordinate of any point represents the sine value, so we will look for the point(s) whose $y\text{-}$coordinate is $0.$

It can be seen that the equation $\sin \theta =0$ has one solution other than $\pi$ in the interval $0\le \theta <2\pi .$ Keep in mind that if we add or subtract a multiple of $2\pi$ radians, the terminal side of the angle will be in the same position. Therefore, the sine of the resulting angles will also be solutions to the equation. Notice that the complete solution of $\sin \theta =0$ does not only include $\pi +2\pi n.$ Therefore, $\sin \theta =0$ is not an example of the type of trigonometric equation that we want to find.

Solution(s) of $\cos \theta =\text{-}1$

Let's now consider $\cos \theta =\text{-}1.$ Since the $x\text{-}$coordinate of any point on the unit circle represents the cosine value, we will look for the point(s) whose $x\text{-}$coordinate is $\text{-}1.$

Note that $P(\text{-}1,0)$ is the only point whose $x\text{-}$coordinate is $\text{-}1.$ This implies that $\pi$ is the only solution to the equation $\cos \theta =\text{-}1$ in the interval $0\le \theta <2\pi .$ Therefore, we can write the complete solution of the equation $\cos \theta =\text{-}1$ as $\pi +2\pi n,$ where $n$ is an integer. This is exactly what we are looking for!

Conclusion

We found that one of the trigonometric equations with the complete solution $\pi +2\pi n$ is $\cos \theta =\text{-}1.$ We can write a second example by multiplying both sides of the Equation I by $2.$ Let's now write a third example by multiplying both sides of the Equation I by $3.$ Note that answers may vary. It is possible to write many different trigonometric equations with the complete solution of $\pi +2\pi n$ by multiplying both sides of the Equation I by any non-zero number.