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  1. Class 11
  2. Important Question for exams Class 11
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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‘₯βˆ’5π‘₯)/2) + sin 3x = 0 2 sin (6π‘₯/2) . cos ((βˆ’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 = – 1 sin 3x = 0 or cos 2x = (βˆ’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 ∈ Z Put y = 0 3x = nΟ€ Β± (-1)n 0 3x = nΟ€ x = π‘›πœ‹/3 where n ∈ Z General solution for cos 2x = (βˆ’πŸ)/𝟐 Let cos x = cos y β‡’ cos 2x = cos 2y Given cos 2x = (βˆ’1)/2 From (3) and (4) cos 2y = (βˆ’1)/2 cos (2y) = cos (2πœ‹/3) β‡’ 2y = 2πœ‹/3 General solution for cos 2x = cos 2y is 2x = 2nΟ€ Β± 2y where n ∈ Z putting 2y = 2πœ‹/3 2x = nΟ€ Β± 2πœ‹/3 x = 1/2 (2nΟ€ Β± 2πœ‹/3) x =2π‘›πœ‹/2 Β± 1/2 Γ— 2πœ‹/3 x = nΟ€ Β± πœ‹/3 where n ∈ Z Hence General Solution is For sin3x = 0, x = π‘›πœ‹/3 and for cos 2x = (βˆ’1)/2 , x = nΟ€ Β± πœ‹/3 where n ∈ Z

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
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