5.3 Solving Trigonometric Equations
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5.3Solving Trigonometric
Equations
JMerrill, 2010
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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 )
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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
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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
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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
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Solve
You Try23tan x 1 0 f or [0,2 ]
3tanx 3
5 7 11x , , ,6 6 6 6
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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
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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.
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Quick review of Identities
Day 2 on 5.3
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Fundamental Trigonometric Identities
Reciprocal Identities
1cscsin
1seccos
1cottan
Also true:
1sincsc
1cossec
1tancot
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Fundamental Trigonometric Identities
Quotient Identities
sintancos
coscotsin
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Fundamental Trigonometric Identities
Pythagorean Identities2 2sin cos 1
2 2tan 1 sec
These are crucial!You MUST know
them.2 21 cot csc
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Pythagorean Memory Trick
sin2 cos2
tan2 cot2
sec2 csc2
(Add the top of the triangle to = the bottom)
1
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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
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Recall:Solving an algebraic equation
2 3 4 0( 1)( 4) 0( 1) 0 ( 4) 0 1 4
x xx xx or xx x
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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
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Solve2sin cos [0 ,360 )o oSolve for
2sin cos cos2sin
2 cot1 tan2
26.6 ,206.6o o
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You cannot divide both sides by a common factor, if the factor cancels out. You will lose a root…
What You CANNOT Do
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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
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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
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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!
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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
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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
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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
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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
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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
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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:
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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
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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
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