# Ex 3.4, 6 - Class 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 3.4, 6 Find the general solution of the equation cos 3x + cos x β cos 2x = 0 cos 3x + cos x β cos 2x = 0 (cos 3x + cos x) β cos 2x = 0 2 cos ((3π₯ + π₯)/2) . cos ((3π₯ β π₯)/2) β cos 2x = 0 2 cos (4π₯/2) . cos (2π₯/2) β cos 2x = 0 2 cos (2x) . cos x β cos 2x = 0 cos (2x) (2cos x β 1) = 0 Hence cos 2x = 0 or (2cos x β 1) = 0 cos 2x = 0 or 2cos x = 1 cos 2x = 0 or cos x = 1/2 We find general solutions of both separately General solution for cos 2x = 0 Let cos x = cos y β cos 2x = cos 2y Given cos 2x = 0 From (1) and (2) cos 2y = 0 cos 2y = cos 90Β° cos 2y = cos π/2 2y = π/2 General solution for cos 2x = cos 2y is given by 2x = 2nΟ Β± 2y where n β Z putting 2y = π/2 2x = 2nΟ Β± π/2 x = 1/2 (2nΟ Β± π/2) x = nΟ Β± π/2 General solution for cos x = π/π Let cos x = cos y Given cos x = 1/2 From (3) and (4) cos y = 1/2 cos (y) = cos (π/3) β y = π/3 General solution for cos x = cos y is x = 2nΟ Β± y where n β Z putting y = π/3 x = nΟ Β± π/3 where n β Z Hence General Solution is For cos 2x = 0, x = nΟ Β± π/2 & for cos x = 1/2 , x = nΟ Β± π/3 where n β Z

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.