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Lesson 5.3 Solving Trigonometric Equations

Lesson 5.3

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Lesson 5.3. Solving Trigonometric Equations. Solving Trigonometric Equations. To solve trigonometric equations:. Use standard algebraic techniques learned in Algebra II. Look for factoring and collecting like terms. Isolate the trig function in the equation. - PowerPoint PPT Presentation

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Page 1: Lesson 5.3

Lesson 5.3

Solving Trigonometric Equations

Page 2: Lesson 5.3

Solving Trigonometric Equations

Use standard algebraic techniques learned in Algebra II.

Look for factoring and collecting like terms.

Isolate the trig function in the equation.

Use the inverse trig functions to assist in determining solutions.

To solve trigonometric equations:

Page 3: Lesson 5.3

Solving Trigonometric Equations

For all problems, The solution interval

Will be[0, 2)

You are responsible for checking your solutions back into the original problem!

Page 4: Lesson 5.3

Solving Trigonometric Equations

2cos 1 0x

2cos 1x

1cos 2x

Solve:

5,3 3

x

Step 1: Isosolate cos x using algebraic skills.

Step 2: Determine in which quadrants cosine is positive. Use the inversefunction to assist by finding the angle in Quad I first. Then use that angleas the reference angle for the other quadrant(s).

QI QIV Note: cosine is positive in Quad I and Quad IV.

Note: The reference angle is /3.

Page 5: Lesson 5.3

Solving Trigonometric Equations

2tan 1 0x 2tan 1x 2tan 1x

Solve:

tan 1x

3 5 7, , ,4 4 4 4

x

Step 1:

Step 2:

Note: Since there is a , all four quadrantshold a solution with /4 being the referenceangle.

Q1 QII QIII QIV

Page 6: Lesson 5.3

Solving Trigonometric Equations

2cot cos 2cotx x x

2cot cos 2cot 0x x x 2cot cos 2 0x x

Solve:

2cot 0 cos 2 0x or x

3,2 2

x

2cos 2x 2cos 2x

cos 2x

x

Step 1:

Step 2: Note: There is no solution here because 2 lies outside the range for cosine.

Page 7: Lesson 5.3

Solving Trigonometric Equations

tan 1 0x

2sec 4 0x

33tan tanx x

Try these:

1.

2.

3.

Solution

5 7 110, , , , ,6 6 6 6

x

3 7,

4 4x

2 4 5, , ,3 3 3 3

x

Page 8: Lesson 5.3

Solving Trigonometric Equations

22sin sin 1 0x x

2sin 1 sin 1 0x x

2sin 1 0 sin 1 0x or x

Solve:

1sin

2x sin 1x

7 11,

6 6x

2x

Factor the quadratic equation.

Set each factor equal to zero.

Solve for sin x

Determine the correct quadrantsfor the solution(s).

Page 9: Lesson 5.3

Solving Trigonometric Equations

22sin 3cos 3 0x x

22 1 cos 3cos 3 0x x 22 2cos 3cos 3 0x x

Solve:

22cos 3cos 1 0x x 22cos 3cos 1 0x x

2cos 1 cos 1 0x x 2cos 1 0 cos 1 0x or x

1cos

2x cos 1x

5,3 3

x

0x

Replace sin2x with 1-cos2x

Distribute

Combine like terms.

Multiply through by – 1.

Factor.Set each factor equal to zero.

Solve for cos x.

Determine the solution(s).

Page 10: Lesson 5.3

Solving Trigonometric Equations

cos 1 sinx x 2 2cos 1 sinx x

2 2cos 2cos 1 sinx x x

Solve:

2 2cos 2cos 1 1 cosx x x 22cos 2cos 0x x

2cos cos 1 0x x 2cos 0 cos 1 0x or x cos 0x cos 1x

3,2 2

x

x

Square both sides of the equationin order to change sine into termsof cosine giving only one trigfunction to work with.

FOIL or Double Distribute

Replace sin2x with 1 – cos2x

Set equation equal to zero since it is a quadratic equation.

Factor

Set each factor equal to zero.Solve for cos x

Determine the solution(s).XWhy is 3/2 removed as a solution? It is removed because it does not

check in the original equation.

Page 11: Lesson 5.3

Solving Trigonometric Equations

1cos3

2x

53 ,

3 3x

5,9 9

x

Solve:

Solution:No algebraic work needs to be done because cosine is already by itself.Remember, 3x refers to an angle and one cannot divide by 3 because it is cos 3x which equals ½.

Since 3x refers to an angle, find the angles whose cosine value is ½.

Now divide by 3 because it is angle equaling angle.

Notice the solutions do not exceed 2. Therefore,more solutions may exist.

Return to the step where you have 3x equalingthe two angles and find coterminal angles for those two.

7 11, ,3 3

53 ,

3 3x

Divide those two new angles by 3.7 11, ,9 9

5,9 9

x

Page 12: Lesson 5.3

Solving Trigonometric Equations

The solutions still do not exceed 2. Return to 3x and find two more coterminal angles.

13 17, ,3 3

7 11, ,3 3

53 ,

3 3x

Divide those two new angles by 3.13 17, ,9 9

7 11, ,9 9

5,9 9

x

The solutions still do not exceed 2. Return to 3x and find two more coterminal angles.

19 23, ,3 3

13 17, ,3 3

7 11, ,3 3

53 ,

3 3x

Divide those two new angles by 3.19,9

13 17, ,9 9

7 11, ,9 9

5,9 9

x

Notice that 19/9 now exceeds 2 andis not part of the solution.

Therefore the solution to cos 3x = ½ is 5 7 11 13 17, , , , ,9 9 9 9 9 9

x

Page 13: Lesson 5.3

Solving Trigonometric Equations

24sin 2cos 1x x

csc cot 1x x

3sin 2

2x

Try these:

2cos

2 2

x

5.4218x

2x

2 5 5 11, , ,

3 6 3 6x

2x

1.

2.

3.

4.

Solution

Page 14: Lesson 5.3

Solving Trigonometric Equations

What you should know:

1. How to use algebraic techniques to solve trigonometric equations.

2. How to solve quadratic trigonometric equations by factoring or the quadratic formula.

3. How to solve trigonometric equations involvingmultiple angles.