1. Chapter 3 Class 11 Trigonometric Functions
2. Serial order wise

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 ((๐ฅ+5๐ฅ)/2) . cos ((๐ฅโ5๐ฅ)/2) + sin 3x = 0 2 sin (6๐ฅ/2) . cos ((โ4๐ฅ)/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 = โ 1 sin 3x = 0 or cos 2x = (โ1)/2 We need to find general solution both separately General solution for sin 3x = 0 Let sin x = sin y โ 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 = nฯ ยฑ (-1)n 3y where n โ Z Put y = 0 3x = nฯ ยฑ (-1)n 0 3x = nฯ x = ๐๐/3 where n โ Z General solution for cos 2x = (โ๐)/๐ Let cos x = cos y โ cos 2x = cos 2y Given cos 2x = (โ1)/2 From (3) and (4) cos 2y = (โ1)/2 cos (2y) = cos (2๐/3) โ 2y = 2๐/3 General solution for cos 2x = cos 2y is 2x = 2nฯ ยฑ 2y where n โ Z putting 2y = 2๐/3 2x = nฯ ยฑ 2๐/3 x = 1/2 (2nฯ ยฑ 2๐/3) x =2๐๐/2 ยฑ 1/2 ร 2๐/3 x = nฯ ยฑ ๐/3 where n โ Z Hence General Solution is For sin3x = 0, x = ๐๐/3 and for cos 2x = (โ1)/2 , x = nฯ ยฑ ๐/3 where n โ Z