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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.