Last updated at May 29, 2018 by Teachoo

Transcript

Example 24 Solve 2 cos2 x + 3 sin x = 0 2 cos2x + 3 sin x = 0 2 (1 − sin2 x) + 3 sin x = 0 2 – 2 sin2x + 3 sin x = 0 – 2sin2x + 3sin x + 2 = 0 Let sin x = a – 2a2 + 3a + 2 = 0 0 = 2a2 – 3a – 2 2a2 – 3a – 2 = 0 2a2 – 4a + a – 2 = 0 2a (a – 2) + 1 (a – 2) = 0 (2a + 1) (a – 2) = 0 Hence 2a + 1 = 0 2a = 0 – 1 2a = – 1 a = (−1)/2 So, a = (−1)/2 & a = 2 Hence sin x = (−1 )/2 or sin x = 2 Value of sin is always between -1 and 1 Hence sin x = 2 is not Possible ∴ sin x = (−1)/2 We find its solution Solving sin x = (−𝟏)/𝟐 Let sin x = sin y given sin x = (−1)/2 From (1) and (2) sin y = (−1)/2 sin y = sin 7𝜋/6 y = 7𝜋/6 General Solution is x = nπ + ( -1 )n y where n ∈ Z Put y = 7𝜋/6 Hence, x = nπ + (-1)n 7𝜋/6 Where n ∈ Z

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Chapter 3 Class 11 Trigonometric Functions

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.