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 45 Page 917

Start by rewriting the equation using the Pythagorean Identity and solving it for Then, draw the unit circle on the coordinate plane.

radians, radians, radians, for any integer

Practice makes perfect
We will start by rewriting the equation using one of the Pythagorean Identities.
After that, we will solve the obtained equation for 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 as Then we will simplify the equation. Let's do it!
Simplify

Solving the Equation

To solve this equation we will introduce a new variable to represent With this change we get a quadratic equation in terms of
In the above quadratic equation, we have that and Let's substitute these values into the Quadratic Formula and simplify.
Simplify right-hand side
To further simplify the equation, we need to separate the right-hand side into two cases.
Since and we found the values of we can rewrite the results in terms of
We obtained two values for Recall that the cosine of an angle in standard position is the coordinate of the point of intersection of its terminal side and the unit circle.
Let's solve the two equations one at a time.

To solve the first equation, we need to consider the points on the unit circle that have an coordinate of

This is the only solution to the first equation on the unit circle. Therefore, all the solutions to the equation are plus a multiple of

To solve the second equation, we need to consider the points on the unit circle that have an coordinate of Recall that cosine is negative in Quadrants II and III.

We can now draw two congruent right triangles, each with a leg on the axis. Since the radius of the unit circle is the hypotenuse of each triangle is also Furthermore, since the coordinate of both points is the length of the leg which is on the axis of each triangle is

For both right triangles, one of the legs is half the hypotenuse, so they are triangles. In this type of triangle, the angle adjacent the shorter leg measures or radians. With this information and knowing that half a turn measures radians, we can calculate the desired angle measures. To do so, we will do two things.

  • Subtract from
  • Add to

Let's do it!

We found two solutions to the given equation, and Finally, 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 given equation.

Solutions to the Given Equation

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