**Ex 3.4, 2**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 3.4, 2 Find the principal and general solutions of the equation sec x = 2 Given sec x = 2 1/cosβ‘π₯ = 2 1/2 = cos x cos x = 1/2 We know that cos 60Β° = 1/2 We find value of x where cos is positive cos is positive in Ist and lVth Quadrant Value in Ist Quadrant = 60Β° Value in lVth Quadrant = 360Β° β 60Β° = 300Β° So Principal solution are x = 60Β° and x = 300Β° x = 60 Γ π/180 and x = 300 Γ 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 β y = π/3 Since cos x = cos y General Solution is x = 2nΟ Β± y where n β Z Put y = π/3 Hence, x = 2nΟ Β± π/3 where n β Z

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.