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Ex 3.4, 2 - sec x = 2, find principal and general solutions - Ex 3.4

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

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Ex 3.4, 2 Find the principal and general solutions of the equation sec x = 2 Given sec x = 2 1/cos⁡𝑥 = 2 1/2 = cos x cos x = 1/2 We know that cos 60° = 1/2 We find value of x where cos is positive cos is positive in Ist and lVth Quadrant Value in Ist Quadrant = 60° Value in lVth Quadrant = 360° – 60° = 300° So Principal solution are x = 60° and x = 300° x = 60 × 𝜋/180 and x = 300 × 𝜋/180 x = 𝜋/3 and x = 5𝜋/3 To find general solution Let cos x = cos y and given cos x = 1/2 From (1) and (2) cos y = 1/2 cos y = cos 𝜋/3 ⇒ y = 𝜋/3 Since cos x = cos y General Solution is x = 2nπ ± y where n ∈ Z Put y = 𝜋/3 Hence, x = 2nπ ± 𝜋/3 where n ∈ Z

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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.