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Ex 3.4, 7 - Find general solution of sin 2x + cos x = 0 - Ex 3.4

Ex 3.4, 7 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 7 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Ex 3.4, 7 - Chapter 3 Class 11 Trigonometric Functions - Part 4


Transcript

Ex 3.4, 7 Find the general solution of the equation sin 2x + cos x = 0 sin 2x + cos x = 0 Putting sin 2x = 2 sin x cos x 2 sin x cos x + cos x = 0 cos x (2sin x + 1) = 0 Hence, We find general solution of both equations separately cos x = 0 2sin x + 1 = 0 2sin x = –1 sin x = (−1)/2 General solution for cos x = 0 Given cos x = 0 General Solution is x = (2n + 1) 𝝅/𝟐 where n ∈ Z General solution for 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 General Solution is x = nπ + (−1)n y where n ∈ Z Putting y = 7𝜋/6 x = nπ + (−1)n 7𝜋/6 Where n ∈ Z Therefore, For cos x = 0, x = (2n + 1) 𝝅/𝟐 OR For sin x = (−1)/2, 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.