Holt Algebra 2 14-6 Solving Trigonometric Equations 14-6 Solving Trigonometric Equations Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation

Holt Algebra 2

14-6 Solving Trigonometric Equations14-6 Solving Trigonometric Equations

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

14-6 Solving Trigonometric Equations

Warm Up

Solve.

1. x2 + 3x – 4 = 0

2. 3x2 + 7x = 6

Evaluate each inverse trigonometric

function.

3. Tan-1 1

4. Sin-1

x = 1 or – 4

45°

– 60

Holt Algebra 2


Solve equations involving trigonometric functions.

Objective

Holt 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 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 +

sinθ =

Subtract sinθ from both sides.

Combine like terms.

Holt Algebra 2


Example 1 Continued

sinθ =

θ = 30°

θ = sin-1

Multiply by 2.

Apply the inverse sineθ.

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.

Use reference angles to find other values of θ.

θ = 150° + 360°n

Holt Algebra 2


Example 1 Continued

1

–1

–90 90

Method 2 Use a graph.

Graph y = sinθ and

y = sinθ + in the same viewing window for –90° ≤ θ ≤ 90°.

Use the intersect feature of your graphing calculator to find the points of intersection.

The graphs intersect at θ = 30°. Thus, θ = 30° + 1360°n, where n is an integer.

Holt 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 –

θ = 150°

Apply the inverse cosineθ.

Find θ when cosine θ = .

Holt Algebra 2


Check It Out! Example 1 Continued

Use reference angles to find other values of θ.

θ = 150° + 360°n, 210° +360°n.

Method 2 Use a graph.

Graph y = 2cosθ and

y = in the same

viewing window for

–360° ≤ θ ≤ 360°.

–360 360

2

–2

The graphs intersect at θ = 150°. Thus, θ = 150° + 360°n, where n is an integer.

Holt Algebra 2


Some trigonometric equations can be solved by applying the same methods used for quadratic equations.

Holt 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 Algebra 2


Example 2A Continued

tanθ = 1 or tan θ =

θ = tan-1(1) θ = tan-1

= 45° or 225° ≈ 36.9° or 216.9°Use a calculator.

Find all angles for 0°≤ θ ≤360°.

Apply the inverse tangent.

Holt Algebra 2


Example 2B: Solving Trigonometric Equations in Quadratic Form

2cos2θ – cosθ = 1 for 0 ≤ θ ≤ .

2cos2θ – cosθ – 1 = 0

(2cosθ + 1) (cosθ – 1) = 0

cosθ = or cosθ = 1

Subtract 1 from both sides.

Factor the quadratic expression by comparing it with 2x2 – x + 1 = 0.

Apply the Zero Product Property.

Find both angles for 0 ≤ θ ≤ .

θ = or θ = 0

Holt Algebra 2


Check It Out! Example 2a

Solve each equation for 0 ≤ θ ≤ 2.

cos2 θ + 2cosθ = 3

cos2 θ + 2cosθ – 3 = 0

(cosθ – 1)(cosθ + 3) = 0

cosθ = 1 or cosθ = –3

cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1.

cosθ = 2 or 0

Subtract 3 from both sides.

Factor the quadratic expression by comparing it to x2 +2x – 3 = 0.


The only solution will come from cosθ = 1.

Holt 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 Algebra 2


Check It Out! Example 2b Continued

Apply the inverse sine.

Use a calculator. Find both angles.

Holt Algebra 2


You can often write trigonometric equations involving more than one function as equations of only one function by using trigonometric identities.

Holt 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

2tan2θ – 2 = 0

tan2θ – 1 = 0

(tanθ – 1)(tanθ + 1) = 0

tanθ = 1 or tanθ = – 1

Substitute 1 + tan2θ for sec2θ by the Pythagorean identity.

Simplify. Divide by 2.

Factor.


Holt Algebra 2


Example 3A Continued

Check Use the intersect feature of your graphing calculator. A graph supports your answer.

Holt 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

–2sin2θ = 0

sin2θ = 0

sinθ = 0

θ = 0° or 180° or 360°

Substitute 1 – sin2θ for cos2θ by the Pythagorean identity.

Simplify.

Divide both sides by – 2.

Take the square root of both sides.

Holt Algebra 2


Example 3B Continued

cos2θ = 1+sin2θ for 0° ≤ θ ≤ 360°

Check Use the intersect feature of your graphing calculator. A graph supports your answer.

θ = 0° or 180° or 360°

Holt Algebra 2


Check It Out! Example 3a

Use trigonometric identities to solve each equation for the given domain.

4sin2θ + 4cosθ = 5

4(1 - cos2θ) + 4cosθ – 5 = 0

4cos2θ – 4cosθ + 1 = 0

(2cos2θ – 1)2 = 0

Substitute 1 – cos2θ for sin2θ by the Pythagorean identity.

Simplify.

Factor.

Take the square root of both sides and simplify.

Holt 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

cosθ(2sinθ + 1) = 0

Substitute 2cosθsinθ for sin2θ by the double-angle identity.

Factor.


Holt 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 Algebra 2


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.

11 Understand the Problem

The answer will be specific dates in the year.

Holt Algebra 2


22 Make a Plan

Substitute 4 for t in the model. Then solve the equation for m by using algebra.

Solve33

4 = 1.665sin (m + 3) + 5.485

sin-1(–0.8918) = (m + 3)

Substitute 4 for t.

Isolate the sine term.

Apply the inverse sine θ.

Holt 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 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 Algebra 2


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.

Look Back44

Check your answer by using a graphing calculator.

Holt 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.



Holt Algebra 2


List the important information:

• The number of hours of sunlight in the day, which is represented by h = 12.

• d represents the number of days after January 1.

• The function model is

(d – 85.25) + 12.22.h(d) = 3.31sin



Holt Algebra 2


22 Make a Plan

Substitute 12 for h in the model. Then solve the equation for d by using algebra.

Solve33

Substitute 12 for h.

Isolate the sine term.

12 = 3.31sin (d – 85.25) + 12.22

Apply the inverse sine θ.

Holt Algebra 2


Sine is negative in Quadrants lll and lV. Compute both values.

Qlll:

81.4 ≈ d

Holt Algebra 2


QlV:

271.6 ≈ d

Holt Algebra 2


Look Back44

Check your answer by using a graphing calculator. Enter

Graph the functions on the same viewing window, and find the points of intersection. The graphs intersect in late March and late September.

y = 3.31sin (d – 85.25) + 12.22

Holt 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.

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Holt Algebra 2 14-6 Solving Trigonometric Equations 14-6 Solving Trigonometric Equations Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation