Exercise 45 Page 917

We will start by rewriting the equation using one of the Pythagorean Identities. After that, we will solve the obtained equation for $\cos \theta .$ Then, we will draw the unit circle on the coordinate plane. Let's do these things one at a time.

Rewriting the Equation

Note that we will need to rearrange the terms of the Pythagorean Identity so that we can use it for our expression. We will start by rewriting ${\sin }^2\theta$ as $1-{\cos }^2\theta .$ Then we will simplify the equation. Let's do it!

Solving the Equation

To solve this equation we will introduce a new variable $z$ to represent $\cos \theta .$ With this change we get a quadratic equation in terms of $z.$ In the above quadratic equation, we have that $a=2,$ $b=\text{-}1,$ and $c=\text{-}1.$ Let's substitute these values into the Quadratic Formula and simplify. To further simplify the equation, we need to separate the right-hand side into two cases.

$z=\frac{1\pm 3}{4}$
$z=\frac{1+3}{4}$	$z=\frac{1-3}{4}$
$z=\frac{4}{4}$	$z=\frac{\text{-}2}{4}$
$z=1$	$z=\text{-}\frac{1}{2}$

Since $z=\cos \theta$ and we found the values of $z,$ we can rewrite the results in terms of $\cos \theta .$ We obtained two values for $\cos \theta .$ Recall that the cosine of an angle in standard position is the $x\text{-}$coordinate of the point of intersection $P$ of its terminal side and the unit circle. Let's solve the two equations one at a time.

$\cos \theta =1$

To solve the first equation, $\cos \theta =1,$ we need to consider the points on the unit circle that have an $x\text{-}$coordinate of $1.$

This is the only solution to the first equation on the unit circle. Therefore, all the solutions to the equation $\cos \theta =1$ are $0$ plus a multiple of $2\pi .$

$\cos \theta =\text{-}\frac{1}{2}$

To solve the second equation, $\cos \theta =\text{-}\frac{1}{2},$ we need to consider the points on the unit circle that have an $x\text{-}$coordinate of $\text{-}\frac{1}{2}.$ Recall that cosine is negative in Quadrants II and III.

We can now draw two congruent right triangles, each with a leg on the $x\text{-}$axis. Since the radius of the unit circle is $1,$ the hypotenuse of each triangle is also $1.$ Furthermore, since the $x\text{-}$coordinate of both points is $\text{-}\frac{1}{2},$ the length of the leg which is on the $x\text{-}$axis of each triangle is $\frac{1}{2}.$

For both right triangles, one of the legs is half the hypotenuse, so they are $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ triangles. In this type of triangle, the angle adjacent the shorter leg measures $60^{\circ }$ or $\frac{\pi }{3}$ radians. With this information and knowing that half a turn measures $\pi$ radians, we can calculate the desired angle measures. To do so, we will do two things.

• Subtract $\frac{\pi }{3}$ from $\pi .$
• Add $\frac{\pi }{3}$ to $\pi .$

Let's do it!

We found two solutions to the given equation, $\theta =\frac{2\pi }{3}$ and $\theta =\frac{4\pi }{3}.$ Finally, 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 given equation.

Solutions to the Given Equation

Let's finally write all the solutions to the given equation. Here, $n$ is any integer.