# Ex 3.4, 7

Last updated at March 6, 2018 by Teachoo

Last updated at March 6, 2018 by Teachoo

Transcript

Ex 3.4, 7 Find the general solution of the equation sin 2x + cos x = 0 sin 2x + cos x = 0 We know that sin 2x = 2 sin x cos x ⇒ 2 sin x cos x + cos x = 0 ⇒ cos x (2sin x + 1) = 0 Hence , cos x = 0 or 2sin x + 1 = 0 cos x = 0 or 2sin x = 0 – 1 cos x = 0 or 2sin x = – 1 cos x = 0 or sin x = (−1)/2 We find general solution of both equations separately General solution for cos x = 0 Let cos x = cos y Given cos x = 0 From (1) and (2) cos y = 0 cos y = cos 𝜋/2 ⇒ y = 𝜋/2 General Solution is x = 2nπ ± y where n ∈ Z Putting y = 𝜋/2 Hence, x = 2nπ ± 𝜋/2 where n ∈ Z Finding general solution for sin x = (−𝟏)/𝟐 Let sin x = sin y given sin x = (−1)/2 From (3) and (4) 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 Hence for cos x = 0, x = 2nπ ± 𝜋/2 where n ∈ Z or for sin x = (−1)/2 x = nπ + (-1)n 7𝜋/6 Where n ∈ Z

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .