Transcript Document

Copyright © 2011 Pearson Education, Inc.
Slide 9.5-1
Chapter 9: Trigonometric Identities and
Equations (I)
9.1 Trigonometric Identities
9.2 Sum and Difference Identities
9.3 Further Identities
9.4 The Inverse Circular Functions
9.5 Trigonometric Equations and Inequalities (I)
9.6 Trigonometric Equations and Inequalities (II)
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-2
9.5 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.
2 sin x  1  0
2 sin x  1
1
sin x 
2
Two values for x, 6 and 56 , satisfy sin x  12 for 0  x  2 .
Therefore the solution set is 6 , 56 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-3
9.5 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 6 and 56 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-4
9.5 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 6 , 56 .
(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 0, 6   56 ,2 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-5
9.5 Solving a Trigonometric Equation
by Factoring
Example Solve sin x tan x = sin x over the interval
[0°, 360°).
Solution
sin x tan x  sin x
sin x tan x  sin x  0
sin x(tan x  1)  0
sin x  0
or
tan x  1  0
tan x  1




x  0 or x  180
x  45 or x  225
The solution set is 0 ,45 ,180 ,225 .




Caution Avoid dividing both sides by sin x. The two solutions
that make sin x = 0 would not appear.
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-6
9.5
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.
tan x  tan x  2  0
(tan x  1)(tan x  2)  0
tan x  1  0
tan x  2  0
or
tan x  1
tan x  2
or
2
The solutions for tan x = 1 in [0, 2) are x = 4 or 54 .
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.
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-7
9.5 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.
 3  9  4  3  13
cot x 

2
2
cot x  3.302775638 or cot x  .3027756377
Since we cannot take the inverse cotangent with the
calculator, we use the fact that cot x  tan1 x .
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-8
9.5 Solving a Trigonometric Equation
Using the Quadratic Formula
cot x  3.302775638
tan x  .3027756377
or
or
cot x  .3027756377
tan x  3.302775638
x  1.276795025
x  .2940013018 or
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}.
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-9
9.5 Solving a Trigonometric Equation by
Squaring and Trigonometric Substitution
Example
Solve tan x  3  sec x over the interval [0, 2).
Solution
Square both sides and use the identity
1 + tan2 x = sec2 x.
tan x  3  sec x
2
2
tan x  2 3 tan x  3  sec x
2
2
tan x  2 3 tan x  3  1  tan x
2 3 tan x  2
1
3
tan x  

3
3
Possible solutions are 56 and 116 . Verify tha t the only
solution is 116 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-10
9.4 The Inverse Sine Function
Solving a Trigonometric Equation Analytically
1. Decide whether the equation is linear or quadratic, so
you can determine the solution method.
2. If only one trigonometric function is present, solve the
equation for that function.
3. If more than one trigonometric function is present,
rearrange the equation so that one side equals 0. Then
try to factor and set each factor equal to 0 to solve.
4. If the equation is quadratic in form, but not factorable,
use the quadratic formula. Check that solutions are in
the desired interval.
5. Try using identities to change the form of the equation.
It may be helpful to square each side of the equation
first. If this is done, check for extraneous values.
Copyright © 2011 Pearson Education, Inc.
Slide 9.5-11
9.4 The Inverse Sine Function
Solving a Trigonometric Equation Graphically
1. For an equation of the form f(x) = g(x), use the
intersection-of-graphs method.
2. For an equation of the form f(x) = 0, use the
x-intercept method.
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Slide 9.5-12