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

Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 4
Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 5

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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‘₯ βˆ’ 5π‘₯)/2) + sin 3x = 0 2 sin (6π‘₯/2) . cos ((βˆ’4π‘₯)/2) + sin 3x = 0 2 sin (3x) . cos (βˆ’2x) + sin 3x = 0 We know that sin x + sin y = 2sin ((π‘₯ + 𝑦)/2) cos ((π‘₯ βˆ’ 𝑦)/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 = –1 cos 2x = (βˆ’1)/2 General solution is 3x = nΟ€ x = (π‘›πœ‹ )/3 where n ∈ Z General solution for cos 2x = (βˆ’πŸ)/𝟐 Let cos x = cos y cos 2x = cos 2y Given cos 2x = (βˆ’1)/2 From (1) and (2) cos 2y = (βˆ’1)/2 cos 2y = (βˆ’1)/2 cos (2y) = cos (2πœ‹/3) 2y = 2πœ‹/3 General solution for cos 2x = cos 2y is 2x = 2nΟ€ Β± 2y Putting 2y = 2πœ‹/3 2x = nΟ€ Β± 2πœ‹/3 Rough We know that cos 60Β° = 1/2 But we need (βˆ’1)/2 So, angle is in 2nd and 3rd quadrant ΞΈ = 60Β° 180 – ΞΈ = 180 – 60 = 120Β° = 120 Γ— πœ‹/180 = 2πœ‹/3 x = 1/2 (2nΟ€ Β± 2πœ‹/3) x = nΟ€ Β± πœ‹/3 where n ∈ Z Hence General Solution is For sin3x = 0, x = 𝒏𝝅/πŸ‘ OR For cos 2x = (βˆ’1)/2 , x = nΟ€ Β± 𝝅/πŸ‘ 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 12 years. He provides courses for Maths and Science at Teachoo.