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Analytic Trigonometry
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2. 7.5 More Trigonometric Equations
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3. Objectives Solving Trigonometric Equations by Using Identities
Equations with Trigonometric Functions of Multiples of Angles

4. More Trigonometric Equations
In this section we solve trigonometric equations by first using identities to simplify the equation. We also solve trigonometric equations in which the terms contain multiples of angles.

5. Solving Trigonometric Equations by Using Identities

6. Solving Trigonometric Equations by Using Identities
In the next example we use trigonometric identities to express a trigonometric equation in a form in which it can be factored.

7. Example 1 – Using a Trigonometric Identity
Solve the equation 1 + sin = 2 cos2.
Solution: We first need to rewrite this equation so that it contains only one trigonometric function. To do this, we use a trigonometric identity:
1 + sin = 2 cos2
1 + sin = 2(1 – sin2 )
2 sin2 + sin – 1 = 0
(2 sin – 1) (sin + 1) = 0
Given equation
Pythagorean identity
Put all terms on one side
Factor

8. Example 1 – Solution 2 sin – 1 = 0 or sin + 1 = 0
cont’d
2 sin – 1 = or sin + 1 = 0
sin = or sin = –1
 = or  =
Because sine has period 2, we get all the solutions of the equation by adding integer multiples of 2 to these solutions.
Set each factor equal to 0
Solve for sin
Solve for sin in interval [0, 2)

9. Example 1 – Solution Thus the solutions are
cont’d
Thus the solutions are
 = k  = k  = k
where k is any integer.

10. Example 3 – Squaring and Using an Identity
Solve the equation cos + 1 = sin in the interval [0, 2).
Solution:
To get an equation that involves either sine only or cosine only, we square both sides and use a Pythagorean identity:
cos + 1 = sin
cos2 + 2 cos + 1 = sin2
cos2 + 2 cos + 1 = 1 – cos2
2 cos2 + 2 cos = 0
Given equation
Square both sides
Pythagorean identity
Simplify

11. Example 3 – Solution 2 cos = 0 or cos + 1 = 0 cos = 0 or cos = –1
cont’d
2 cos = or cos + 1 = 0
cos = or cos = –1
 = or  = 
Because we squared both sides, we need to check for extraneous solutions. From Check Your Answers we see that the solutions of the given equation are  /2 and .
Set each factor equal to 0
Solve for cos
Solve for  in [0, 2)

12. Example 3 – Solution Check Your Answer:  =  =  = 
cont’d
Check Your Answer:
 =  =  = 
cos = sin cos = sin cos  + 1 = sin 
0 + 1 = ≟ – –1 + 1 = 0

13. Example 4 – Finding Intersection Points
Find the values of x for which the graphs of f (x) = sin x and g (x) = cos x intersect.
Solution 1: Graphical The graphs intersect where f (x) = g (x). In Figure 1 we graph y1 = sin x and y2 = cos x on the same screen, for x between 0 and 2.
(a)
(b)
Figure 1

14. Example 4 – Solution 1
cont’d
Using or the intersect command on the graphing calculator, we see that the two points of intersection in this interval occur where x  and x 
Since sine and cosine are periodic with period 2, the intersection points occur where
x  k and x  k
where k is any integer.

15. Example 4 – Solution 2 Algebraic
cont’d
Algebraic
To find the exact solution, we set f (x) = g (x) and solve the resulting equation algebraically:
sin x = cos x
Since the numbers x for which cos x = 0 are not solutions of the equation, we can divide both sides by cos x:
= 1
tan x = 1
Equate functions
Divide by cos x
Reciprocal identity

16. Example 4 – Solution 2
cont’d
The only solution of this equation in the interval (– /2,  /2) is x =  /4. Since tangent has period  , we get all solutions of the equation by adding integer multiples of  :
x = k
where k is any integer. The graphs intersect for these values of x.
You should use your calculator to check that, rounded to three decimals, these are the same values that we obtained in Solution 1.

17. Equations with Trigonometric Functions of Multiples of Angles

18. Equations with Trigonometric Functions of Multiples of Angles
When solving trigonometric equations that involve functions of multiples of angles, we first solve for the multiple of the angle, then divide to solve for the angle.

19. Example 5 – A Trigonometric Equation Involving Multiple of an Angle
Consider the equation 2 sin 3 – 1 = 0.
(a) Find all solutions of the equation.
(b) Find the solutions in the interval [0, 2).
Solution:
(a) We first isolate sin 3 and then solve for the angle 3.
2 sin 3 – 1 = 0
2 sin 3 = 1
sin 3 =
Given equation
Add 1
Divide by 2

20. Example 5 – Solution 3 = cont’d
3 =
Solve for 3 in the interval [0, 2) (see Figure 2)
Figure 2

21. Example 5 – Solution
cont’d
To get all solutions, we add integer multiples of 2 to these solutions. So the solutions are of the form
3 = k  = k
To solve for , we divide by 3 to get the solutions
where k is any integer.

22. Example 5 – Solution
cont’d
(b) The solutions from part (a) that are in the interval [0, 2) correspond to k = 0, 1, and 2. For all other values of k the corresponding values of  lie outside this interval.
So the solutions in the interval [0, 2) are

23. Example 6 – A Trigonometric Equation Involving a Half Angle
Consider the equation
(a) Find all solutions of the equation.
(b) Find the solutions in the interval [0, 4).
Solution:
(a) We start by isolating tan :
Given equation
Add 1

24. Example 6 – Solution
cont’d
Since tangent has period , to get all solutions we add integer multiples  of to this solution. So the solutions are of the form
Divide by
Solve for in the interval

25. Example 6 – Solution Multiplying by 2, we get the solutions  = + 2k
cont’d
Multiplying by 2, we get the solutions
 = k
where k is any integer.
(b) The solutions from part (a) that are in the interval [0, 4) correspond to k = 0 and k = 1. For all other values of k the corresponding values of x lie outside this interval. Thus the solutions in the interval [0, 4) are