Principal and General Solutions

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Question 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 ((đĨ + 5đĨ)/2) . cos ((đĨ â 5đĨ)/2) + sin 3x = 0 2 sin (6đĨ/2) . cos ((â4đĨ)/2) + sin 3x = 0 2 sin (3x) . cos (â2x) + sin 3x = 0 We know that sin x + sin y = 2sin ((đĨ + đĻ)/2) cos ((đĨ â đĻ)/2) Replacing x by x & y by 5x 2 sin 3x . cos 2x + sin 3x = 0 sin 3x (2cos 2x + 1) = 0 Hence We need to find general solution both separately General solution for sin 3x = 0 Given sin 3x = 0 sin 3x = 0 2cos 2x + 1 = 0 2cos 2x = â1 cos 2x = (â1)/2 General solution is 3x = nĪ x = (đđ )/3 where n â Z General solution for cos 2x = (âđ)/đ Let cos x = cos y cos 2x = cos 2y Given cos 2x = (â1)/2 From (1) and (2) cos 2y = (â1)/2 cos 2y = (â1)/2 cos (2y) = cos (2đ/3) 2y = 2đ/3 General solution for cos 2x = cos 2y is 2x = 2nĪ Âą 2y Putting 2y = 2đ/3 2x = nĪ Âą 2đ/3 Rough We know that cos 60Â° = 1/2 But we need (â1)/2 So, angle is in 2nd and 3rd quadrant Î¸ = 60Â° 180 â Î¸ = 180 â 60 = 120Â° = 120 Ã đ/180 = 2đ/3 x = 1/2 (2nĪ Âą 2đ/3) x = nĪ Âą đ/3 where n â Z Hence General Solution is For sin3x = 0, x = đđ/đ OR For cos 2x = (â1)/2 , x = nĪ Âą đ/đ where n â Z