Explanation Calculation: Solve the equation sin 4 x − sin 2 x = 0 . sin 4 x − sin 2 x = 0 2 sin 2 x cos 2 x − sin 2 x = 0 sin 2 x ( 2 cos 2 x − 1 ) = 0 sin 2 x = 0 or 2 cos 2 x − 1 = 0 Note that, the sines of angles is positive in first and second quadrant and cosines of angles is positive in first and fourth quadrant. 2 x = sin − 1 ( 0 ) or 2 cos 2 x = 1 2 x = sin − 1 ( 0 ) or 2 x = cos − 1 ( 1 2 ) 2 x = 0 , π , 2 π or 2 x = π 3 , 2 π − π 3 , 2 π + π 3 , 4 π − π 3

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 20.5, Problem 12E

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To solve: The trigonometric equation sin4x−sin2x=0 for all values of x such that 0≤x<2π.

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Chapter 20.5, Problem 11E

Chapter 20.5, Problem 13E

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Chapter 20 Solutions

Basic Technical Mathematics

Ch. 20.1 - Prob. 1PE Ch. 20.1 - Prob. 2PE Ch. 20.1 - Prob. 3PE Ch. 20.1 - Prob. 1E Ch. 20.1 - Prob. 2E Ch. 20.1 - In Exercises 36, use a calculator to check the...Ch. 20.1 - Prob. 4E Ch. 20.1 - Prob. 5E Ch. 20.1 - Prob. 6E Ch. 20.1 - Prob. 7E

Ch. 20.1 - Prob. 8E Ch. 20.1 - Prob. 9E Ch. 20.1 - Prob. 10E Ch. 20.1 - Prob. 11E Ch. 20.1 - Prob. 12E Ch. 20.1 - Prob. 13E Ch. 20.1 - Prob. 14E Ch. 20.1 - Prob. 15E Ch. 20.1 - Prob. 16E Ch. 20.1 - Prob. 17E Ch. 20.1 - Prob. 18E Ch. 20.1 - In Exercises 19–38, prove the given identities. Ch. 20.1 - Prob. 20E Ch. 20.1 - In Exercises 19–38, prove the given identities. 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The trigonometric equation sin 4 x − sin 2 x = 0 for all values of x such that 0 ≤ x &lt; 2 π .

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To solve: The trigonometric equation sin4x−sin2x=0 for all values of x such that 0≤x<2π.

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Chapter 20 Solutions

The trigonometric equation sin 4 x − sin 2 x = 0 for all values of x such that 0 ≤ x < 2 π .