copyright © 2007 pearson education, inc. slide 9-1 6.2 trigonometric equations and inequalities (i)...
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Copyright © 2007 Pearson Education, Inc. Slide 9-1
6.2 Trigonometric Equations and Inequalities (I)
• Solving a Trigonometric Equation by Linear Methods
Example Solve 2 sin x – 1 = 0 over the interval [0, 2).
Analytic Solution Since this equation involves the first power of sin x, it is linear in sin x.
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sin
1sin2
01sin2
x
x
x
., isset solution theTherefore
.20for sinsatisfy , and ,for valuesTwo
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21
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xxx
Copyright © 2007 Pearson Education, Inc. Slide 9-2
6.2 Solving a Trigonometric Equation by Linear Methods
Graphing Calculator Solution
Graph y = 2 sin x – 1 over the interval [0, 2].
The x-intercepts have the same decimal approximations as . and 6
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Copyright © 2007 Pearson Education, Inc. Slide 9-3
6.2 Solving Trigonometric Inequalities
Example Solve for x over the interval [0, 2).
(a) 2 sin x –1 > 0 and (b) 2 sin x –1 < 0.
Solution
(a) Identify the values for which the graph of y = 2 sin x –1 is above the x-axis. From the previous graph, the solution set is
(b) Identify the values for which the graph of y = 2 sin x –1 is below the x-axis. From the previous graph, the solution set is
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Copyright © 2007 Pearson Education, Inc. Slide 9-4
6.2 Equations Solvable by Factoring
Example Solve tan2 x + tan x –2 = 0 over the interval [0, 2).
Solution This equation is quadratic in term tan x.
The solutions for tan x = 1 in [0, 2) are x = Use a calculator to find the solution to tan-1(–2) –1.107148718. To get the values in the interval [0, 2), we add and 2 to tan-1(–2) to get
x = tan-1(–2) + 2.03443936 and x = tan-1(–2) + 2 5.176036589.
0)2)(tan1(tan
02tantan2
xx
xx
1tan
01tan
x
x
2tan
02tan
x
x
or
or
. 45
4 or
Copyright © 2007 Pearson Education, Inc. Slide 9-5
6.2 Solving a Trigonometric Equation by Factoring
Example Solve sin x tan x = sin x over the interval [0°, 360°).
Solution
0)1(tansin0sintansinsintansin
xxxxx
xxx
0sin x1tan01tanor
xx
225or45180or0 xxxx
.225,180,45,0 isset solution The
Caution Avoid dividing both sides by sin x. The two solutions that make sin x = 0 would not appear.
Copyright © 2007 Pearson Education, Inc. Slide 9-6
6.2 Solving a Trigonometric Equation Using the Quadratic Formula
Example Solve cot x(cot x + 3) = 1 over the interval [0, 2).
Solution Rewrite the expression in standard quadratic
form to get cot2 x + 3 cot x – 1 = 0, with a = 1, b = 3, c = –1, and cot x as the variable.
Since we cannot take the inverse cotangent with the calculator, we use the fact that
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cotx
3027756377.cotor302775638.3cot xx
.cot tan1
xx
Copyright © 2007 Pearson Education, Inc. Slide 9-7
6.2 Solving a Trigonometric Equation Using the Quadratic Formula
The first of these, –.29400113018, is not in the desired interval. Since the period of cotangent is , we add and then 2 to –.29400113018 to get 2.847591352 and 5.989184005.
The second value, 1.276795025, is in the interval, so we add to it to get another solution.
The solution set is {1.28, 2.85, 4.42, 5.99}.
3027756377.cotor302775638.3cot xx
2940013018.x or 276795025.1x
302775638.3tanor3027756377.tan xx
Copyright © 2007 Pearson Education, Inc. Slide 9-8
6.2 Solving a Trigonometric Equation by Squaring and Trigonometric Substitution
Example Solve over the interval [0, 2).
Solution Square both sides and use the identity
1 + tan2 x = sec2 x.
xx sec3tan
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tan
2tan32
tan13tan32tan
sec3tan32tan
sec3tan
22
22
x
x
xxx
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only t theVerify tha . and are solutions Possible
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