Copyright © Cengage Learning. All rights reserved. CHAPTER Equations 6.

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Copyright © Cengage Learning. All rights reserved. CHAPTER Equations 6

Copyright © Cengage Learning. All rights reserved. Trigonometric Equations Involving Multiple Angles SECTION 6.3

3 Solve a simple trigonometric equation involving a multiple angle. Use an identity to solve a trigonometric equation involving a multiple angle. Solve a trigonometric equation involving a multiple angle by factoring or the quadratic formula. Solve a real-life problem using a trigonometric equation. 1 Learning Objectives 2 4 3

4 Example 1 Solve, if 0    < 360 . Solution: The equation cannot be simplified further. The reference angle is given by Because the cosine function is positive in QI or QIV, the expression 2  must be coterminal with 30  or 330 . Therefore, for any integer k, all solutions will have the form

5 Example 1 – Solution Because we were asked only for values of  between 0  and 360 , we substitute the appropriate integers in place of k. For all other values of k,  will fall outside the given interval. cont’d

6 Example 3 Solve sin 2x cos x + cos 2x sin x =, if 0  x < 2 . Solution: We can simplify the left side by using the formula for sin (A + B).

7 Example 3 – Solution First we find all possible solutions for x: cont’d

8 Example 3 – Solution To find those solutions that lie in the interval 0  x < 2 , we let k take on values of 0, 1, and 2. Doing so results in the following solutions (Figure 5): cont’d Figure 5

9 Example 4 Find all solutions to 2 sin 2 3  – sin 3  – 1 = 0, if  is measured in degrees. Solution: We have an equation that is quadratic in sin 3 . We factor and solve as usual.

10 Example 4 – Solution or or cont’d

11 Example 7 Table 3 shows the average monthly attendance at Lake Nacimiento in California. Table 3

12 Example 7 The average monthly attendance at Lake Nacimiento could be modeled by the function where x is the month, with x = 1 corresponding to January. Use the model to determine the percentage of the year that the average attendance is at least 25,000. cont’d

13 Example 7 – Solution To begin, we will find the values of x for which y is exactly 25,000. To do so, we must solve the equation First we isolate the trigonometric function.

14 Example 7 – Solution The reference angle is cos –1 (0.2992)  1.27 radians. The cosine is positive in QI and QIV, so we have The solution x = is outside the given interval 0  x  12. However, because the period is 12, adding or subtracting any multiple of 12 to will give another solution. cont’d

15 Example 7 – Solution Thus, x = – 12 = 2.57 is a valid solution within the given interval. Looking at Figure 7, we see that y will be at least 25,000 between x = 2.57 and x = 7.43, inclusive. This gives us or cont’d Figure 7