Exercise 54 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 $\cos \theta .$ To find a solution for this equation, we will use an inverse trigonometric function and a calculator. One solution to $\cos \theta =\text{-}\frac{1}{2}$ is $\theta =\frac{2\pi }{3}.$ To find the second solution for this equation, we need to use the unit circle. Remember, cosine is negative in the second and third quadrants of the coordinate plane.

The cosine of an angle in standard position is the $x\text{-}$coordinate of the point of intersection of the unit circle and the terminal side of the angle. The solution $\theta =\frac{2\pi }{3}$ is in Quadrant II, so the second solution must be in Quadrant III. Because these angles are symmetric across the $x\text{-}$axis and a full turn measures $2\pi$ radians, to find this angle we subtract $\frac{2\pi }{3}$ from $2\pi .$

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 equation. These are the $x\text{-}$intercepts of the graph of $y=2\cos \theta +1.$