Section 7.5 Solving Trigonometric Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education,...
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Transcript of Section 7.5 Solving Trigonometric Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education,...
![Page 1: Section 7.5 Solving Trigonometric Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.](https://reader035.fdocuments.net/reader035/viewer/2022062322/5697bfd91a28abf838caf8c2/html5/thumbnails/1.jpg)
Section 7.5
Solving Trigonometric
Equations
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
![Page 2: Section 7.5 Solving Trigonometric Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.](https://reader035.fdocuments.net/reader035/viewer/2022062322/5697bfd91a28abf838caf8c2/html5/thumbnails/2.jpg)
Objectives
Solve trigonometric equations.
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Solve Trigonometric Equations
When an equation contains a trigonometric expression with a variable, such as cos x, it is called a trigonometric equation. Some trigonometric equations are identities, such as sin2 x + cos2 x = 1. Now we consider equations, such as 2 cos x = –1, that are usually not identities. As we have done for other types of equations, we will solve such equations by finding all values for x that make the equation true.
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Example
Solve 2cos x 1.
First solve for cos x.
2π/3 and 4π/3 have cosine –1/2.
Solution:
23
2k or 43
2k
cos x 1
2
These numbers, plus any multiple of 2π are the solutions.
120ºk 360º or 240ºk 360º
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Example
Solve 3tan2x 3 in the interval 0,2 .
First solve for tan 2x.
We are looking for solutions x to the equation for which
0 ≤ x < 2π.
Solution:tan2x 1
Multiplying by 2, we get 0 ≤ 2x < 4πwhich is the interval we use when solving tan 2x = –1.
Using the unit circle, find the points 2x in [0, 4π) for which tan 2x = –1.
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They are:
Example (CONT)
The values of x, are found by dividing each of these by 2.
2x 34
,74
,11
4,15
4
x 38
,78
,11
8,15
8
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Example
Solve 1
cos 1 1.2108 in [0º ,360º ).2
Solution: 1
2cos 1 1.2108
Use a calculator in DEGREE mode to find the reference angle = cos–1 0.4216 ≈ 65.06º.
1
2cos 0.2108
cos 0.4216
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Example (cont)
Since cos is positive, the solutions are in quadrants I and IV. The solutions in [0º, 360º) are
65.06ºand360º – 65.06º = 294.94º
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Example
Solve 2cos2 u 1 cosu in 0º ,360º .
Use the principal of zero products:
Solution: 2cos2 u 1 cosu
The solutions in [0º, 360º) are 60º, 180º and 300º.
2cos2 u cosu 1 0
2cosu 1 cosu 1 0
2cosu 1 0 or cosu 1 02cosu 1 or cosu 1
cosu 1
2 or cosu 1
u 60º ,300º or u 180º
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Graphical Solution: INTERSECT METHOD
Example (cont)
and use the INTERSECT feature on the calculator.
y1 2cos2 x and y2 1 cos xGraph the equations:
The left most solution is 60º. Use the INTERSECT feature two more times to find the solutions, 180º and 300º.
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Example (cont)
Graphical Solution: ZERO METHOD
2cos2 x cos x 1 0Write the equation in the form
The left most zero is 60º. Use the ZERO feature two more times to find the solutions, 180º and 300º.
Then graphy 2cos2 x cos x 1
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Example
Solve 10sin2 x 12sin x 7 0 in 0º ,360º .
Use the quadratic formula: a = 10, b = –12, and c = –7.Solution:
No solution.
sin x 12 12 2 4 10 7
210
sin x 12 144 280
20
12 424
20
12 20.5913
20sin x 1.6296 or sin x 0.4296
reference angle: 25.44º
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Example (cont)
Sin x is negative, the solutions are in quadrants III & IV.
The solutions are180º + 25.44º = 205.44ºand300º – 25.44º = 334.56º.
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Example
Solve the following in [0, 2π).
Solution:
sin cos cotx x x
We cannot find the solutions algebraically. We can approximate them with a graphing calculator.
On the screen on the next slide, on the left side we use the INTERSECT METHOD. Graph
y1 sin x cos x and y2 cot x
On the screen on the right side we use the ZERO METHOD. Graph
y1 sin x cos x cot x
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Example (cont)
The solutions in [0, 2π) are approximately 1.13 and 5.66.