Download 1 Trigonometric Equations Using Factoring

Trigonometric Equations Using
Factoring
Bradley Hughes
Larry Ottman
Lori Jordan
Mara Landers
Andrea Hayes
Brenda Meery
Art Fortgang
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Printed: August 3, 2015
AUTHORS
Bradley Hughes
Larry Ottman
Lori Jordan
Mara Landers
Andrea Hayes
Brenda Meery
Art Fortgang
www.ck12.org
C HAPTER
Chapter 1. Trigonometric Equations Using Factoring
1
Trigonometric Equations
Using Factoring
Here you’ll learn how to factor trig equations and then solve them using the factored form.
Solving trig equations is an important process in mathematics. Quite often you’ll see powers of trigonometric
functions and be asked to solve for the values of the variable which make the equation true. For example, suppose
you were given the trig equation
2 sin x cos x = cos x
Could you solve this equation? (You might be tempted to just divide both sides by cos x, but that would be incorrect
because you would lose some solutions.) Instead, you’re going to have to use factoring. Read this Concept, and at
its conclusion, you’ll be ready to factor the above equation and solve it.
Watch This
MEDIA
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James Sousa Example: Solve a Trig Equation by Factoring
Guidance
You have no doubt had experience with factoring. You have probably factored equations when looking for the
possible values of some variable, such as "x". It might interest you to find out that you can use the same factoring
method for more than just a variable that is a number. You can factor trigonometric equations to find the possible
values the function can take to satisfy an equation.
Algebraic skills like factoring and substitution that are used to solve various equations are very useful when solving
trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by zero during these
maneuvers.
Example A
Solve 2 sin2 x − 3 sin x + 1 = 0 for 0 < x ≤ 2π.
Solution:
1
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2 sin2 x − 3 sin x + 1 = 0
Factor this like a quadratic equation
(2 sin x − 1)(sin x − 1) = 0
↓
&
2 sin x − 1 = 0
or sin x − 1 = 0
2 sin x = 1
1
sin x =
2
5π
π
x = and x =
6
6
sin x = 1
π
x=
2
Example B
Solve 2 tan x sin x + 2 sin x = tan x + 1 for all values of x.
Solution:
Pull out sin x
There is a common factor of (tan x + 1)
Think of the −(tan x + 1) as (−1)(tan x + 1), which is why there is a −1 behind the 2 sin x.
Example C
Solve 2 sin2 x + 3 sin x − 2 = 0 for all x, [0, π].
Solution:
2 sin2 x + 3 sin x − 2 = 0 → Factor like a quadratic
(2 sin x − 1)(sin x + 2) = 0
.
&
2 sin x − 1 = 0
sin x + 2 = 0
1
sin x =
sin x = −2
2
π
5π
x = and x =
There is no solution because the range of sin x is [−1, 1].
6
6
2
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Chapter 1. Trigonometric Equations Using Factoring
Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will
result in a true expression.
Guided Practice
1. Solve the trigonometric equation 4 sin x cos x + 2 cos x − 2 sin x − 1 = 0 such that 0 ≤ x < 2π.
2. Solve tan2 x = 3 tan x for x over [0, π].
3. Find all the solutions for the trigonometric equation 2 sin2 4x − 3 cos 4x = 0 over the interval [0, 2π).
Solutions:
1. Use factoring by grouping.
2 sin x + 1 = 0
or
2 sin x = −1
1
sin x = −
2
7π 11π
x= ,
6 6
2 cos x − 1 = 0
2 cos x = 1
1
cos x =
2
π 5π
x= ,
3 3
2.
tan2 x = 3 tan x
tan2 x − 3 tan x = 0
tan x(tan x − 3) = 0
tan x = 0
x = 0, π
or
tan x = 3
x = 1.25
3.
2 sin2 4x − 3 cos 4x = 0
3
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x
x
2 1 − cos2
− 3 cos = 0
4
4
x
x
2 − 2 cos2 − 3 cos = 0
4
4
x
x
2 cos2 + 3 cos − 2 = 0
4 4 x
x
2 cos − 1 cos + 2 = 0
4
4
.
&
x
x
2 cos − 1 = 0 or cos + 2 = 0
4
4
x
x
2 cos = 1
cos = −2
4
4
x 1
cos =
4 2
5π
x π
=
or
4 3
3
4π
20π
x=
or
3
3
20π
3
is eliminated as a solution because it is outside of the range and cos 4x = −2 will not generate any solutions
because −2 is outside of the range of cosine. Therefore, the only solution is 4π
3 .
Concept Problem Solution
The equation you were given is
2 sin x cos x = cos x
To solve this:
2 sin x cos x = cos x
Subtract cos x from both sides and factor it out of the equation:
2 sin x cos x − cos x = 0
cos x(2 sin x − 1) = 0
Now set each factor equal to zero and solve. The first is cos x:
cos x = 0
π 3π
x= ,
2 2
And now for the other term:
4
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Chapter 1. Trigonometric Equations Using Factoring
2 sin x − 1 = 0
1
sin x =
2
π 5π
x= ,
6 6
Explore More
Solve each equation for x over the interval [0, 2π).
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
cos2 (x) + 2 cos(x) + 1 = 0
1 − 2 sin(x) + sin2 (x) = 0
2 cos(x) sin(x) − cos(x) = 0
sin(x) tan2 (x) − sin(x) = 0
sec2 (x) = 4
sin2 (x) − 2 sin(x) = 0
3 sin(x) = 2 cos2 (x)
2 sin2 (x) + 3 sin(x) = 2
tan(x) sin2 (x) = tan(x)
2 sin2 (x) + sin(x) = 1
2 cos(x) tan(x) − tan(x) = 0
sin2 (x) + sin(x) = 2
tan(x)(2 cos2 (x) + 3 cos(x) − 2) = 0
sin2 (x) + 1 = 2 sin(x)
2 cos2 (x) − 3 cos(x) = 2
5