5.3 Solving Trigonometric Equations

5.3Solving Trigonometric

Equations

JMerrill, 2010

It will be imperative that you know the identities from Section 5.1. Concentrate on the reciprocal, quotient, and Pythagorean identities.

The Pythagorean identities are crucial!

Recall (or Relearn )

Solve Using the Unit Circle Solve sin x = ½ Where on the circle does the sin x = ½ ?

5,6 6x

52 , 26 6x n n

Particular Solutions

General solutions

Solve for [0,2π]

Find all solutions

Solving a Trigonometric Equation Using Algebra

22sin 1 0 [0 ,360 )o oSolve for 22sin 1 0

22sin 1 2 1sin

2

2sin2

4 sin

2 2 .

There are solutionsbecause is positivein quadrants andnegative in quadrants

45 ,135 ,225 ,315o o o o

Find all solutions to: sin x + = -sin x

Using Algebra Again2

sinx sinx 2 0

2sinx 2

2sinx 2

5 7x 2n and x 2n4 4

Solve

You Try23tan x 1 0 f or [0,2 ]

3tanx 3

5 7 11x , , ,6 6 6 6

Solve by Factoring

sin tan 3 [0, )n 2six x xSolve for

sin tan 3sin 0x x x sin (tan 3) 0x x sin 0 tan 3x or x

Round to nearest hundredth

0,3.14 1.25,4.39x x

Solve

You Try2cot xcos x 2cot x in [0,2 )

2cot xcos x 2cot x 0 2cot x(cos x 2) 0

cot x 03x ,2 2

2

2

cos x 2 0cos x 2cosx 2DNE (Does Not Exist)No solution

Verify graphically

These 2 solutions are true because of the interval specified. If we did not specify and interval, you answer would be based on the period of tan x which is π and your only answer would be the first answer.

Quick review of Identities

Day 2 on 5.3

Fundamental Trigonometric Identities

Reciprocal Identities

1cscsin

1seccos

1cottan

Also true:

1sincsc

1cossec

1tancot

Fundamental Trigonometric Identities

Quotient Identities

sintancos

coscotsin

Fundamental Trigonometric Identities

Pythagorean Identities2 2sin cos 1

2 2tan 1 sec

These are crucial!You MUST know

them.2 21 cot csc

Pythagorean Memory Trick

sin2 cos2

tan2 cot2

sec2 csc2

(Add the top of the triangle to = the bottom)

1

Sometimes You Must Simplify Before you Can Solve

Strategies Change all functions to sine and cosine (or at

least into the same function) Substitute using Pythagorean Identities Combine terms into a single fraction with a

common denominator Split up one term into 2 fractions Multiply by a trig expression equal to 1 Factor out a common factor

Recall:Solving an algebraic equation

2 3 4 0( 1)( 4) 0( 1) 0 ( 4) 0 1 4

x xx xx or xx x

Solve

2 2sin sin cosx x x Hint: Make the words match so use a Pythagorean identity2 2sin sin 1 sinx x x

Quadratic: Set = 02 2sin sin 1 sin 0x x x Combine like

terms22sin sin 1 0x x Factor—(same as 2x2-x-

1)(2sin 1)(sin 1) 0x x

1sin sin 12

x or x 7 11, ,

2 6 6x

Solve2sin cos [0 ,360 )o oSolve for

2sin cos cos2sin

2 cot1 tan2

26.6 ,206.6o o

You cannot divide both sides by a common factor, if the factor cancels out. You will lose a root…

What You CANNOT Do

Example

2sin cos cos2sin

2 cot1 tan2

sin tan 3sinsin tan 3sin

sin sin tan 3

x x xx x xx xx

Common factor—lost a root

No common factor = OK

Sometimes, you must square both sides of an equation to obtain a quadratic. However, you must check your solutions. This method will sometimes result in extraneous solutions.

Squaring and Converting to a Quadratic

Solve cos x + 1 = sin x in [0, 2π) There is nothing you can do. So, square

both sides (cos x + 1)2 = sin2x cos2x + 2cosx + 1 = 1 – cos2x 2cos2x + 2cosx = 0 Now what?

Squaring and Converting to a Quadratic

Remember—you want the words to match so use a Pythagorean substitution!

2cos2x + 2cosx = 0 2cosx(cosx + 1) = 0 2cosx = 0 cosx + 1 = 0 cosx = 0 cosx = -1

Squaring and Converting to a Quadratic

3, 2 2

x x

3, 2 2

x x

Check Solutions

cos 1 sin2 2

0 1 1

3 3cos 1 sin2 2

0 1 1

cos 1 sin1 1 0

Solve 2cos3x – 1 = 0 for [0,2π) 2cos3x = 1 cos3x = ½ Hint: pretend the 3 is not there and solve

cosx = ½ . Answer:

But….

Functions With Multiple Angles

1 1cos2

5,3 3

x

x

Functions With Multiple Angles In our problem 2cos3x – 1 = 0 What is the 2? What is the 3? This graph is happening 3 times as often as

the original graph. Therefore, how many answers should you have?

amplitudefrequency

6

Functions With Multiple Angles

1 1cos2

5,3 3

x

x

Add a whole circle to each of these 7 11,3 3

And add the circle once again.

13 17,3 3

Functions With Multiple Angles

5 7 11 13 173 , , , , ,3 3 3 3 3

5 7 11 13 17, , , , ,9 9 9

,9 9

3

9

x

So x

Final step: Remember we pretended the 3 wasn’t there, but since it is there, x is really 3x:

Work the problems by yourself. Then compare answers with someone sitting next to you.

Round answers: 1. csc x = -5 (degrees)

2. 2 tanx + 3 = 0 (radians)

3. 2sec2x + tanx = 5 (radians)

Practice Problems

o o191.5 ,348.5

2.16, 5.30

2.16, 5.30, .79, 3.93

4. 3sinx – 2 = 5sinx – 1

5. cos x tan x = cos x

6. cos2 - 3 sin = 3

Practice – Exact Answers Only (Radians) Compare Answers

7 11,6 6

3 5, , ,2 2 4 4

32