web analytics

Ex 3.4, 9 - Find general solution of sin x + sin 3x + sin 5x = 0 - Finding general solutions

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

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‘ฅโˆ’5๐‘ฅ)/2) + sin 3x = 0 2 sin (6๐‘ฅ/2) . cos ((โˆ’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 = (โˆ’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 โˆˆ Z Put y = 0 3x = nฯ€ ยฑ (-1)n 0 3x = nฯ€ x = ๐‘›๐œ‹/3 where n โˆˆ Z General solution for cos 2x = (โˆ’๐Ÿ)/๐Ÿ Let cos x = cos y โ‡’ cos 2x = cos 2y Given cos 2x = (โˆ’1)/2 From (3) and (4) cos 2y = (โˆ’1)/2 cos (2y) = cos (2๐œ‹/3) โ‡’ 2y = 2๐œ‹/3 General solution for cos 2x = cos 2y is 2x = 2nฯ€ ยฑ 2y where n โˆˆ Z putting 2y = 2๐œ‹/3 2x = nฯ€ ยฑ 2๐œ‹/3 x = 1/2 (2nฯ€ ยฑ 2๐œ‹/3) x =2๐‘›๐œ‹/2 ยฑ 1/2 ร— 2๐œ‹/3 x = nฯ€ ยฑ ๐œ‹/3 where n โˆˆ Z Hence General Solution is For sin3x = 0, x = ๐‘›๐œ‹/3 and for cos 2x = (โˆ’1)/2 , x = nฯ€ ยฑ ๐œ‹/3 where n โˆˆ Z

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
Jail