Solving Trigonometric Equations

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Sharp Tutor Mon, Jan 17, 2022 09:05 PM UTC

Our objective is to solve equations involving trigonometric functions.

1. 11-6Solving
11-6 SolvingTrigonometric
TrigonometricEquations
Equations
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Holt Algebra 2Algebra 2

2. 11-6 Solving Trigonometric Equations
Warm Up
Solve.
1. x2 + 3x – 4 = 0 x = 1 or – 4
2. 3x2 + 7x = 6
Evaluate each inverse trigonometric
function.
3. Tan-1 1 45°
4. Sin-1 – 60
Holt McDougal Algebra 2

3. 11-6 Solving Trigonometric Equations
Objective
Solve equations involving trigonometric functions.
Holt McDougal Algebra 2

4. 11-6 Solving Trigonometric Equations
Unlike trigonometric identities, most trigonometric
equations are true only for certain values of the
variable, called solutions. To solve trigonometric
equations, apply the same methods used for solving
algebraic equations.
Holt McDougal Algebra 2

5. 11-6 Solving Trigonometric Equations
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.
Holt McDougal Algebra 2

6. 11-6 Solving Trigonometric Equations
Example 1 Continued
sinθ = Multiply by 2.
θ = sin-1 Apply the inverse sineθ.
θ = 30° Find θ when sinθ =
Find all real number value of θ, where n is an
integer.
θ = 30° + 360°n Use the period of the sine function.
θ = 150° + 360°n Use reference angles to find
other values of θ.
Holt McDougal Algebra 2

7. 11-6 Solving Trigonometric Equations
Example 1 Continued
Method 2 Use a graph.
1
Graph y = sinθ and
y = sinθ + in the
same viewing window –90 90
for –90° ≤ θ ≤ 90°.
Use the intersect feature of
your graphing calculator to –1
find the points of
intersection.
The graphs intersect at θ = 30°. Thus,
θ = 30° + 1360°n, where n is an integer.
Holt McDougal Algebra 2

8. 11-6 Solving Trigonometric Equations
Check It Out! Example 1
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.
θ = 150° Find θ when cosine θ = .
Holt McDougal Algebra 2

9. 11-6 Solving Trigonometric Equations
Check It Out! Example 1 Continued
θ = 150° + 360°n, Use reference angles to find
210° +360°n. other values of θ.
Method 2 Use a graph. 2
Graph y = 2cosθ and
y= in the same –360
360
viewing window for
–360° ≤ θ ≤ 360°.
–2
The graphs intersect at θ = 150°.
Thus, θ = 150° + 360°n, where n is an integer.
Holt McDougal Algebra 2

10. 11-6 Solving Trigonometric Equations
Some trigonometric equations can be solved
by applying the same methods used for
quadratic equations.
Holt McDougal Algebra 2

11. 11-6 Solving Trigonometric Equations
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.
Holt McDougal Algebra 2

12. 11-6 Solving Trigonometric Equations
Example 2A Continued
tanθ = 1 or tan θ =
Apply the inverse
θ = tan-1(1) θ = tan-1 tangent.
Use a calculator.
= 45° or 225° ≈ 36.9° or 216.9° Find all angles
for 0°≤ θ ≤360°.
Holt McDougal Algebra 2

13. 11-6 Solving Trigonometric Equations
Example 2B: Solving Trigonometric Equations in
Quadratic Form
2cos2θ – cosθ = 1 for 0 ≤ θ ≤ .
2cos2θ – cosθ – 1 = 0 Subtract 1 from both sides.
(2cosθ + 1) (cosθ – 1) = 0 Factor the quadratic
expression by comparing it
with 2x2 – x + 1 = 0.
cosθ = or cosθ = 1 Apply the Zero Product
Property.
θ= or θ = 0 Find both angles for
0 ≤ θ ≤ .
Holt McDougal Algebra 2

14. 11-6 Solving Trigonometric Equations
Check It Out! Example 2a
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
(cosθ – 1)(cosθ + 3) = 0
to x2 +2x – 3 = 0.
cosθ = 1 or cosθ = –3 Apply the Zero Product
Property.
cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1.
cosθ = 2 or 0 The only solution will come
from cosθ = 1.
Holt McDougal Algebra 2

15. 11-6 Solving Trigonometric Equations
Check It Out! 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 =
Holt McDougal Algebra 2

16. 11-6 Solving Trigonometric Equations
Check It Out! Example 2b Continued
Apply the inverse sine.
Use a calculator. Find both
angles.
Holt McDougal Algebra 2

17. 11-6 Solving Trigonometric Equations
You can often write trigonometric equations
involving more than one function as equations of
only one function by using trigonometric identities.
Holt McDougal Algebra 2

18. 11-6 Solving Trigonometric Equations
Example 3A: Solving Trigonometric Equations with
Trigonometric Identities
Use trigonometric identities to solve each
equation.
tan2θ + sec2θ = 3 for 0 ≤ θ ≤ 2π.
tan2θ + (1 + tan2θ) – 3 = 0 Substitute 1 + tan θ for sec θ
2 2
by the Pythagorean
2tan2θ – 2 = 0 identity.
tan2θ – 1 = 0 Simplify. Divide by 2.
(tanθ – 1)(tanθ + 1) = 0 Factor.
tanθ = 1 or tanθ = – 1 Apply the Zero Product
Property.
Holt McDougal Algebra 2

19. 11-6 Solving Trigonometric Equations
Example 3A Continued
Check Use the intersect feature of your graphing
calculator. A graph supports your answer.
Holt McDougal Algebra 2

20. 11-6 Solving Trigonometric Equations
Example 3B: Solving Trigonometric Equations with
Trigonometric Identities
Use trigonometric identities to solve each equation.
cos2θ = 1 + sin2θ for 0° ≤ θ ≤ 360°
(1 – sin2θ) – 1– sin2θ = 0 Substitute 1 – sin2θ for cos2θ
by the Pythagorean identity.
–2sin2θ = 0 Simplify.
sin2θ = 0 Divide both sides by – 2.
sinθ = 0 Take the square root of both
sides.
θ = 0° or 180° or 360°
Holt McDougal Algebra 2

21. 11-6 Solving Trigonometric Equations
Example 3B Continued
cos2θ = 1+sin2θ for 0° ≤ θ ≤ 360°
θ = 0° or 180° or 360°
Check Use the intersect feature of your graphing
calculator. A graph supports your answer.
Holt McDougal Algebra 2

22. 11-6 Solving Trigonometric Equations
Check It Out! Example 3a
Use trigonometric identities to solve each
equation for the given domain.
4sin2θ + 4cosθ = 5
Substitute 1 – cos 2
θ for sin 2
θ
4(1 - cos θ) + 4cosθ – 5 = 0 by the Pythagorean identity.
2
4cos2θ – 4cosθ + 1 = 0 Simplify.
(2cos2θ – 1)2 = 0 Factor.
Take the square root of both
sides and simplify.
Holt McDougal Algebra 2

23. 11-6 Solving Trigonometric Equations
Check It Out! Example 3b
Use trigonometric identities to solve each
equation for the given domain.
sin2θ = – cosθ for 0 ≤ θ < 2
2cosθsinθ + cosθ = 0 Substitute 2cosθsinθ for sin2θ
by the double-angle identity.
cosθ(2sinθ + 1) = 0 Factor.
Apply the Zero Product
Property.
Holt McDougal Algebra 2

24. 11-6 Solving Trigonometric Equations
Example 4: Problem-Solving Application
On what days does the sun rise at
4 A.M. on Cadillac Mountain? The time
of the sunrise can be modeled by
Holt McDougal Algebra 2

25. 11-6 Solving Trigonometric Equations
1 Understand the Problem
The answer will be specific dates in the year.
List the important information:
• The function model is
t(m) = 1.665 (m + 3) + 5.485.
• Sunrise is at 4 A.M., which is represented
by t = 4.
• m represents the number of months after
January 1.
Holt McDougal Algebra 2

26. 11-6 Solving Trigonometric Equations
2 Make a Plan
Substitute 4 for t in the model. Then solve
the equation for m by using algebra.
3 Solve
4 = 1.665sin (m + 3) + 5.485 Substitute 4 for t.
Isolate the sine
term.
sin-1(–0.8918) = (m + 3) Apply the inverse
sine θ.
Holt McDougal Algebra 2

27. 11-6 Solving Trigonometric Equations
Sine is negative in Quadrants lll and lV.
Compute both values.
Qlll: π + sin-1(0.8918) QlV: 2π + sin-1(0.8918)
Holt McDougal Algebra 2

28. 11-6 Solving Trigonometric Equations
Using an average of 30 days per month, the
date m = 5.10 corresponds to June 4(5 months
and 3 days after January 1) and m = 6.90
corresponds to July 28 (6 months and 27 days
after January 1).
Holt McDougal Algebra 2

29. 11-6 Solving Trigonometric Equations
4 Look Back
Check your answer by using a graphing
calculator.
Enter
y = 1.665sin (x + 3) + 5.485 and y = 4.
Graph the functions on the same viewing
window, and find the points of intersection.
The graphs intersect at early June and late
July.
Holt McDougal Algebra 2

30. 11-6 Solving Trigonometric Equations
Check It Out! Example 4
The number of hours h of sunlight in a day
at Cadillac Mountain can be modeled by
h(d) = 3.31sin (d – 85.25) + 12.22,
where d is the number of days after January
1. When are there 12 hours of sunlight.
1 Understand the Problem
The answer will be specific dates in the year.
Holt McDougal Algebra 2

31. 11-6 Solving Trigonometric Equations
1 Understand the Problem
The answer will be specific dates in the year.
List the important information:
• The function model is
h(d) = 3.31sin (d – 85.25) + 12.22.
• The number of hours of sunlight in the
day, which is represented by h = 12.
• d represents the number of days after
January 1.
Holt McDougal Algebra 2

32. 11-6 Solving Trigonometric Equations
2 Make a Plan
Substitute 12 for h in the model. Then solve
the equation for d by using algebra.
3 Solve
12 = 3.31sin (d – 85.25) + 12.22
Substitute 12 for h.
Isolate the sine term.
Apply the inverse sine θ.
Holt McDougal Algebra 2

33. 11-6 Solving Trigonometric Equations
Sine is negative in Quadrants lll and lV.
Compute both values.
Qlll:
81.4 ≈ d
Holt McDougal Algebra 2

34. 11-6 Solving Trigonometric Equations
QlV:
271.6 ≈ d
Holt McDougal Algebra 2

35. 11-6 Solving Trigonometric Equations
4 Look Back
Check your answer by using a graphing
calculator.
Enter
y = 3.31sin (d – 85.25) + 12.22
Graph the functions on the same viewing
window, and find the points of intersection.
The graphs intersect in late March and late
September.
Holt McDougal Algebra 2

36. 11-6 Solving Trigonometric Equations
Lesson Quiz
1. Find all solutions for cosθ = – cosθ.
θ = 45° + n  360° or 315° + n  360°
2. Solve 3sin2θ – 4 = 0 for 0 ≤ θ ≤ 360°.
θ ≈ 221.8° or 318.2°
3. Solve cos2θ = 3sinθ + 2 for 0 ≤ 0 ≤ 2π.
Holt McDougal Algebra 2

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