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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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