1. Evaluate each inverse trigonometric function.
Warm Up
Solve.
1. x2 + 3x – 4 = 0
2. 3x2 + 7x = 6
Evaluate each inverse trigonometric function.
3. sin-1 1
4. Sin-1
x = 1 or – 4
90°
– 60

2. Objective
Solve equations involving trigonometric functions.

3. Example 1: Solving Trigonometric Equations with Infinitely Many Solutions
Find all the solutions of sinθ = sinθ +
Method 1 Use algebra.
Solve for θ over the principal value of sine, –90° ≤ θ ≤ 90°.
sinθ = sinθ +
sinθ sinθ =
Subtract sinθ from both sides.
sinθ =
Combine like terms.

4. Example 1 Continued
sinθ =
Multiply by 2.
θ = sin-1
Apply the inverse sineθ.
Find θ when sinθ =
θ = 30°
Find all real number value of θ, where n is an integer.
θ = 30° + 360°n
Use the period of the sine function.
Use reference angles to find other values of θ.
θ = 150° + 360°n

5. Example 1 Continued
Method 2 Use a graph. 1 –1
–90
Graph y = sinθ and
y = sinθ + in the same viewing window for –90° ≤ θ ≤ 90°. 90
Use the intersect feature of your graphing calculator to find the points of intersection.
The graphs intersect at θ = 30°. Thus,
θ = 30° °n, where n is an integer.

6. Find all of the solutions of 2cosθ + = 0.
Method 1 Use algebra.
Solve for θ over the principal value of sine, 0 ≤ θ ≤ .
2cosθ =
Subtract from both sides.
cosθ =
Divide both sides by 2.
θ = cos-1 –
Apply the inverse cosineθ.
Find θ when cosine θ =
θ = 150°
θ = 150° + 360°n,
210° +360°n.

7. Example 2A: Solving Trigonometric Equations in Quadratic Form
Solve each equation for the given domain.
4tan2θ – 7 tanθ + 3 = 0 for 0° ≤ θ ≤ 360°.
4tan2θ – 7 tanθ + 3 = 0
Factor the quadratic expression by comparing it with
4x2 – 7x + 3 = 0.
(tanθ – 1)(4tanθ – 3) = 0
Apply the Zero Product Property.

8. Example 2A Continued
tanθ = 1 or tan θ =
Apply the inverse tangent.
θ = tan-1(1) θ = tan-1
Use a calculator. Find all angles for 0°≤ θ ≤360°.
= 45° or 225°
≈ 36.9° or 216.9°

9. Example 2B: Solving Trigonometric Equations in Quadratic Form
2cos2θ – cosθ = 1 for 0 ≤ θ ≤ .
2cos2θ – cosθ – 1 = 0
Subtract 1 from both sides.
Factor the quadratic expression by comparing it with 2x2 – x + 1 = 0.
(2cosθ + 1) (cosθ – 1) = 0
cosθ = or cosθ = 1
Apply the Zero Product Property.
θ = or θ = 0
Find both angles for 0 ≤ θ ≤ .

10. Solve each equation for 0 ≤ θ ≤ 2.
cos2 θ + 2cosθ = 3
Subtract 3 from both sides.
cos2 θ + 2cosθ – 3 = 0
Factor the quadratic expression by comparing it to x2 +2x – 3 = 0.
(cosθ – 1)(cosθ + 3) = 0
Apply the Zero Product Property.
cosθ = 1 or cosθ = –3
cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1.
The only solution will come from cosθ = 1.
cosθ = 2 or 0

11. Solve each equation for 0 ≤ θ ≤ 2.
Example 2b
Solve each equation for 0 ≤ θ ≤ 2.
sin2θ + 5 sinθ – 2 = 0
The equation is in quadratic form but can not be easily factored. Use the quadratic formula.
sinθ =
Apply the inverse sine.
Use a calculator. Find both angles.

12. 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
The solutions are
 = k  = k
where k is any integer.

13. (b) We start by isolating tan  : tan2  – 3 = 0 tan2  = 3
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.

14. 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 

15. 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

16. The second equation has no solution because cos  is never greater than 1.
Thus the solutions are
 = k  = k
where k is any integer.

17. 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 

18. 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)

19. 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

20. 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.

21. HW 28 page 396 #1, 5-15 odd, odd, 26, 29, 30