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Example 24 - Solve 2 cos^2 x + 3 sin x = 0 - General Solution -Teachoo

Example 24 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 24 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Example 24 - Chapter 3 Class 11 Trigonometric Functions - Part 4 Example 24 - Chapter 3 Class 11 Trigonometric Functions - Part 5


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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 So, our equation becomes sin2 x + cos2 x = 1 cos2 x = 1 – sin2 x –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 = –1 a = (−1)/2 a – 2 = 0 a = 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 Rough We know that sin 30° = 1/2 But we need (−1)/2 So, angle is in 3rd & 4th quadrant θ = 30° 180 + θ = 180 + 30 = 210° = 210 × 𝜋/180 = 7/6 π Rough We know that sin 30° = 1/2 But we need (−1)/2 So, angle is in 3rd & 4th quadrant θ = 30° 180 + θ = 180 + 30 = 210° = 210 × 𝜋/180 = 7/6 π General Solution is x = nπ + (−1)n y where n ∈ Z Putting y = 7𝜋/6 x = nπ + (−1)n 𝟕𝝅/𝟔 Where n ∈ Z

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.