1. Class 11
2. Important Question for exams Class 11

Transcript

Ex 3.4, 9 Find the general solution of the equation sin x + sin3x + sin5x = 0 sin x + sin 3x + sin 5x = 0 (sin x + sin 5x) + sin 3x =0 (sin x + sin 5x) + sin 3x = 0 2 sin ((π₯+5π₯)/2) . cos ((π₯β5π₯)/2) + sin 3x = 0 2 sin (6π₯/2) . cos ((β4π₯)/2) + sin 3x = 0 2 sin (3x) . cos (-2x) + sin 3x = 0 2 sin 3x . cos 2x + sin 3x = 0 sin 3x (2cos 2x + 1) = 0 Hence sin 3x = 0 or 2cos 2x + 1 = 0 sin 3x = 0 or 2cos 2x = β 1 sin 3x = 0 or cos 2x = (β1)/2 We need to find general solution both separately General solution for sin 3x = 0 Let sin x = sin y β sin 3x = sin 3y Given sin 3x = 0 From (1) and (2) sin 3y = 0 sin 3y = sin (0) 3y = 0 y = 0 General solution for sin 3x = sin 3y is 3x = nΟ Β± (-1)n 3y where n β Z Put y = 0 3x = nΟ Β± (-1)n 0 3x = nΟ x = ππ/3 where n β Z General solution for cos 2x = (βπ)/π Let cos x = cos y β cos 2x = cos 2y Given cos 2x = (β1)/2 From (3) and (4) cos 2y = (β1)/2 cos (2y) = cos (2π/3) β 2y = 2π/3 General solution for cos 2x = cos 2y is 2x = 2nΟ Β± 2y where n β Z putting 2y = 2π/3 2x = nΟ Β± 2π/3 x = 1/2 (2nΟ Β± 2π/3) x =2ππ/2 Β± 1/2 Γ 2π/3 x = nΟ Β± π/3 where n β Z Hence General Solution is For sin3x = 0, x = ππ/3 and for cos 2x = (β1)/2 , x = nΟ Β± π/3 where n β Z

Class 11
Important Question for exams Class 11