Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
2. Solving Trigonometric Equations Using Inverses
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Exercise 58 Page 917

The intercepts of the graph of a function occur when Let's substitute for into the given equation.
We will start by solving the equation for To do so, we will factor out 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.
We obtained, in total, three equations.
Let's solve these equations one at a time.

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

These points correspond to angles of and radians. These are the solutions to the equation

To find a solution to this equation, we will use an inverse trigonometric function and a calculator.

One solution to is 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 coordinate of the point of intersection of the angle's terminal side and the unit circle. The solution is in Quadrant I, so the second solution must be in Quadrant II. Because these angles are symmetric across the axis and half a turn measures radians, to find this angle we subtract from

Another solution to the equation is

To find a solution to this equation, we will use an inverse trigonometric function and a calculator.

One solution to is 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 is in Quadrant IV, so the second solution must be in Quadrant III, and these angles are symmetric across the axis. Knowing that half a turn measures radians and a full turn measures radians, we can calculate the desired angle measures. We will add to and subtract from

The solutions to the equation are and

intercepts

We found six solutions to the given equation, and Keep in mind that if we add or subtract a multiple of 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 solutions differ by and is of Also, the other solutions differ from each other by and is a of With this information in mind, we can write our answer in a simpler form.
These are the intercepts of the given function.