Exercise 58 Page 917

The $x\text{-}$intercepts of the graph of a function occur when $y=0.$ Let's substitute $0$ for $y$ into the given equation. We will start by solving the equation for $\sin \theta .$ To do so, we will factor out ${\sin }^2\theta$ and use the Zero Product Property. Now that the first equation is fully simplified, let's take a closer look at the second equation. To further simplify Equation II, we need to separate the right-hand side into the positive and negative case.

$\sin \theta =\pm \sqrt{\frac{1}{2}}$
$\sin \theta =\sqrt{\frac{1}{2}}$	$\sin \theta =\text{-}\sqrt{\frac{1}{2}}$
$\sin \theta =\sqrt{\frac{2}{4}}$	$\sin \theta =\text{-}\sqrt{\frac{2}{4}}$
$\sin \theta =\frac{\sqrt{2}}{2}$	$\sin \theta =\text{-}\frac{\sqrt{2}}{2}$

We obtained, in total, three equations. Let's solve these equations one at a time.

$\sin \theta =0$

Recall that the sine of an angle in standard position is the second coordinate of the point of intersection of the angle's terminal side and the unit circle. There are two points on the unit circle with second coordinate $0.$

These points correspond to angles of $\theta =0$ and $\theta =\pi$ radians. These are the solutions to the equation $\sin \theta =0.$

$\sin \theta =\frac{\sqrt{2}}{2}$

To find a solution to this equation, we will use an inverse trigonometric function and a calculator. One solution to $\sin \theta =\frac{\sqrt{2}}{2}$ is $\theta =\frac{\pi }{4}.$ To find a second solution to this equation, we need to use the unit circle. Remember, sine is positive in the first and second quadrants of the coordinate plane.

The sine of an angle in standard position is the $y\text{-}$coordinate of the point of intersection of the angle's terminal side and the unit circle. The solution $\theta =\frac{\pi }{4}$ is in Quadrant I, so the second solution must be in Quadrant II. Because these angles are symmetric across the $y\text{-}$axis and half a turn measures $\pi$ radians, to find this angle we subtract $\frac{\pi }{4}$ from $\pi .$

Another solution to the equation $\sin \theta =\frac{\sqrt{2}}{2}$ is $\theta =\frac{3\pi }{4}.$

$\sin \theta =\text{-}\frac{\sqrt{2}}{2}$

To find a solution to this equation, we will use an inverse trigonometric function and a calculator. One solution to $\sin \theta =\text{-}\frac{\sqrt{2}}{2}$ is $\theta =\text{-}\frac{\pi }{4}.$ To find a second solution to this equation, we need to use the unit circle. Remember, sine is negative in the third and fourth quadrants of the coordinate plane.

The solution $\theta =\text{-}\frac{\pi }{4}$ is in Quadrant IV, so the second solution must be in Quadrant III, and these angles are symmetric across the $y\text{-}$axis. Knowing that half a turn measures $\pi$ radians and a full turn measures $2\pi$ radians, we can calculate the desired angle measures. We will add $\frac{\pi }{4}$ to $\pi ,$ and subtract $\frac{\pi }{4}$ from $2\pi .$

The solutions to the equation $\sin \theta =\text{-}\frac{\sqrt{2}}{2}$ are $\theta =\frac{5\pi }{4}$ and $\theta =\frac{7\pi }{4}.$

$x\text{-}$intercepts

We found six solutions to the given equation, $\theta =0,$ $\theta =\pi ,$ $\theta =\frac{\pi }{4},$ $\theta =\frac{3\pi }{4},$ $\theta =\frac{5\pi }{4},$ and $\theta =\frac{7\pi }{4}.$ 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 resulting angles will also be solutions to the equation. Note that the first $\text{two}$ solutions differ by $\pi ,$ and $\pi$ is $\text{half}$ of $2\pi .$ Also, the other $\text{four}$ solutions differ from each other by $\frac{\pi }{2},$ and $\frac{\pi }{2}$ is a $\text{quarter}$ of $2\pi .$ With this information in mind, we can write our answer in a simpler form. These are the $x\text{-}$intercepts of the given function.