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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
OH nooooooo!
DAY 48 March 16/17, 2015
Class Opener: Find D and R:
a. y= 5x2+2x­3
b. y= 3x­2
c. y= 1 +2
x­3
d. 2x+5=10
Nov 14­2:45 PM
Learning Goal 5.1.-5.2. QUIZ
Trigonometric Identities.
Mar 13­11:56 AM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
IN YOUR NOTES
LESSON 5.3. SOLVING TRIGONOMETRIC EQUATIONS
Determine whether the following equations are true or false:
1.] 2sin(y) = sin(2y)
2.] cos2(y) = cos(y2)
3.] 3tan(x2) ­ 5tan(x2) = ­ 2 tan (x2)
4.] cos(3y) + cos(y) = cos (4y)
5.] cos(y) + cos(x) = cos (x+y)
6.] sin2(3x) = (sin (3x))2
Feb 14­2:54 PM
LESSON 5.3. SOLVING TRIGONOMETRIC EQUATIONS
1.] 2sin(y) = sin(2y)
False
2.] cos2(y) = cos(y2)
False
3.] 3tan(x2) ­ 5tan(x2) = ­ 2 tan (x2) True
4.] cos(3y) + cos(y) = cos (4y) False
5.] cos(y) + cos(x) = cos (x+y) False
6.] sin2(3x) = (sin (3x))2 True
Feb 14­2:54 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Lesson 5.3. Solving Trigonometric Equations (Easy cases)
When solving a trigonometric equation, your goal is to isolate the trigonometric function involved in the equation using standard algebraic operations and trigonometric identities. When solving algebraic equations, we can always CHECK (code word for verify) solutions.
Example: Verify that each x value is a solution of the equation given:
1.] 2 cos(x) – 1 = 0 (a) x = π/3 (b) x = 5π/3
2.] sec x – 2 = 0 (a) x = π/3 (b) x = 5π/3
Feb 14­7:51 PM
Verifying solutions for Trig Equation:
Example: Verify that each x value is a solution of the equation given: Note: solution to the equation is
"zero" of the equation
1.] 2 cos(x) – 1 = 0 (a) x = π/3 (b) x = 5π/3
2.] sec x – 2 = 0 (a) x = π/3 (b) x = 5π/3
Feb 14­10:43 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Lesson 5.3. Solving Trigonometric Equations (Easy cases)
When solving a trigonometric equation, your goal is to isolate the trigonometric function involved in the equation using standard algebraic operations and trigonometric identities. When solving algebraic equations, we can always CHECK (code word for verify) solutions.
Example 1: Solve 2sinx ­ 1 = 0 for x.
Feb 28­3:14 PM
Lesson 5.3. Solving Trigonometric Equations (Easy cases)
When solving a trigonometric equation, your goal is to isolate the trigonometric function involved in the equation using standard algebraic operations and trigonometric identities. When solving algebraic equations, we can always CHECK (code word for verify) solutions.
Example 1: Solve 2sinx ­ 1 = 0 for x.
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term. therefore The sine function is positive in quadrants I and II.
Therefore, two of the solutions to the problem are and Numerical Check: • Left Side:
• Right Side: 0 Feb 28­3:14 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Solve 2sinx ­ 1 = 0 for x.
Therefore, two of the solutions to the problem are and These two solutions are solution on the interval [0, 2π], but there are infitinely many solutions id the domain is not restricted to a specific interval. There will be 2 solution in every 2π revolutions. If you rotate from π/6 exactly 2π radians, you will hit your next solution. If you rotate from 5π/6 exactly 2π radians, you will hit your next solution. Therefore, ALL GENERAL
SOLUTIONS WILL BE
where n is the number of revolutions in positive/negative direction.
Feb 28­3:29 PM
Solve for x:
General solutions: Feb 16­2:11 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Note: General solutions for tangent
repeat every π radians!
Feb 16­2:14 PM
Feb 16­2:19 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Feb 16­2:28 PM
General Solutions:
Feb 17­3:26 PM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Rewriting as a Single Trigonometric Function:
Feb 16­8:41 AM
Remember to use ± when taking the square root!!!
Feb 17­11:11 AM
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5.3. DAY 48 Solving Trig Equations.notebook
March 17, 2015
Feb 14­10:55 PM
Lesson 5.3. 364/ 2­30 EVEN
An Answer Key posted on Mrs. Ghillany's LMSA Web Page
Feb 16­9:38 AM
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