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Analytic Trigonometry
Copyright © Cengage Learning. All rights reserved.

2. 7.4 Basic Trigonometric Equations
Copyright © Cengage Learning. All rights reserved.

3. Objectives Basic Trigonometric Equations
Solving Trigonometric Equations by Factoring

4. Basic Trigonometric Equations

5. Basic Trigonometric Equations
An equation that contains trigonometric functions is called a trigonometric equation. For example, the following are trigonometric equations:
sin2  + cos2  = sin – 1 = tan 2 – 1 = 0
The first equation is an identity—that is, it is true for every value of the variable . The other two equations are true only for certain values of .
To solve a trigonometric equation, we find all the values of the variable that make the equation true.

6. Basic Trigonometric Equations
Solving any trigonometric equation always reduces to solving a basic trigonometric equation—an equation of the form T( ) = c, where T is a trigonometric function and c is a constant.
In the next examples we solve such basic equations.

7. Example 1 – Solving a Basic Trigonometric Equation
Solve the equation
Solution: Find the solutions in one period. Because sine has period 2, we first find the solutions in any interval of length 2. To find these solutions, we look at the unit circle in Figure 1.
Figure 1

8. Example 1 – Solution
cont’d
We see that sin  = in Quadrants I and II, so the solutions
in the interval [0, 2) are
Find all solutions. Because the sine function repeats its values every 2 units, we get all solutions of the equation by adding integer multiples of 2 to these solutions:
where k is any integer.

9. Example 1 – Solution
cont’d
Figure 2 gives a graphical representation of the solutions.
Figure 2

10. Example 4 – Solving a Basic Trigonometric Equation
Solve the equation tan  = 2.
Solution: Find the solutions in one period. We first find one solution by taking tan–1 of each side of the equation.
tan  = 2
 = tan–1(2)
  1.12
Given equation
Take tan–1 of each side
Calculator (in radian mode)

11. Example 4 – Solution
cont’d
By the definition of tan–1 the solution that we obtained is the only solution in the interval (– /2,  /2) (which is an interval of length ).
Find all solutions. Since tangent has period , we get all solutions of the equation by adding integer multiples of  :
  k
where k is any integer.

12. Example 4 – Solution
cont’d
A graphical representation of the solutions is shown in Figure 6.
You can check that the solutions shown in the graph correspond to k = –1, 0, 1, 2, 3.
Figure 6

13. Basic Trigonometric Equations
In the next example we solve trigonometric equations that are algebraically equivalent to basic trigonometric equations.

14. Example 5 – Solving Trigonometric Equations
Find all solutions of the equation.
(a) 2 sin  – 1 = (b) tan2  – 3 = 0
Solution: (a) We start by isolating sin  .
2 sin  – 1 = 0
2 sin  = 1
sin  =
Given equation
Add 1
Divide by 2

15. Example 5 – Solution
cont’d
This last equation is the same as that in Example 1. The solutions are
 = k  = k
where k is any integer.
(b) We start by isolating tan .
tan2  – 3 = 0
tan2  = 3
tan  =
Given equation
Add 3
Take the square root

16. Example 5 – Solution
cont’d
Because tangent has period , we first find the solutions in any interval of length . In the interval (– /2,  /2) the solutions are  =  /3 and  = – /3.
To get all solutions, we add integer multiples of  to these solutions:
 = k  = – k
where k is any integer.

17. Solving Trigonometric Equations by Factoring

18. Solving Trigonometric Equations by Factoring
Factoring is one of the most useful techniques for solving equations, including trigonometric equations.
The idea is to move all terms to one side of the equation, factor, and then use the Zero-Product Property.

19. Example 6 – A Trigonometric Equation of Quadratic Type
Solve the equation 2 cos2  – 7 cos  + 3 = 0.
Solution:
We factor the left-hand side of the equation.
2 cos2  – 7 cos  + 3 = 0
(2 cos  – 1)(cos  – 3) = 0
2 cos  – 1 = 0 or cos  – 3 = 0
cos  = or cos  = 3
Given equation
Factor
Set each factor equal to 0
Solve for cos 

20. Example 6 – Solution
cont’d
Because cosine has period 2, we first find the solutions in the interval [0, 2). For the first equation the solutions are  =  /3 and  = 5 /3 (see Figure 7).
Figure 7

21. Example 6 – Solution
cont’d
The second equation has no solution because cos  is never greater than 1.
Thus the solutions are
 = k  = k
where k is any integer.

22. Example 7 – Solving a Trigonometric Equation by Factoring
Solve the equation 5 sin  cos  + 4 cos  = 0.
Solution: We factor the left-hand side of the equation.
5 sin  cos  + 2 cos  = 0
cos  (5 sin  + 2) = 0
cos  = 0 or 5 sin  + 4 = 0
sin  = –0.8
Given equation
Factor
Set each factor equal to 0
Solve for sin 

23. Example 7 – Solution
cont’d
Because sine and cosine have period 2, we first find the solutions of these equations in an interval of length 2.
For the first equation the solutions in the interval [0, 2) are  =  /2 and  = 3 /2 . To solve the second equation, we take sin–1 of each side.
sin  = –0.80
 = sin–1(–0.80)
Second equation
Take sin–1 of each side

24. Example 7 – Solution
cont’d
  –0.93
So the solutions in an interval of length 2 are  = –0.93 and  =   4.07 (see Figure 8).
Calculator (in radian mode)
Figure 8

25. Example 7 – Solution
cont’d
We get all the solutions of the equation by adding integer multiples of 2 to these solutions.
 = k  = k
  – k   k
where k is any integer.