Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 17, 2020 by Teachoo
Ex 3.4
Ex 3.4, 2 Deleted for CBSE Board 2022 Exams
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Ex 3.4, 9 Important Deleted for CBSE Board 2022 Exams You are here
Ex 3.4
Last updated at Feb. 17, 2020 by Teachoo
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₯ β 5π₯)/2) + sin 3x = 0 2 sin (6π₯/2) . cos ((β4π₯)/2) + sin 3x = 0 2 sin (3x) . cos (β2x) + sin 3x = 0 We know that sin x + sin y = 2sin ((π₯ + π¦)/2) cos ((π₯ β π¦)/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 = β1 cos 2x = (β1)/2 General solution is 3x = nΟ x = (ππ )/3 where n β Z General solution for cos 2x = (βπ)/π Let cos x = cos y cos 2x = cos 2y Given cos 2x = (β1)/2 From (1) and (2) cos 2y = (β1)/2 cos 2y = (β1)/2 cos (2y) = cos (2π/3) 2y = 2π/3 General solution for cos 2x = cos 2y is 2x = 2nΟ Β± 2y Putting 2y = 2π/3 2x = nΟ Β± 2π/3 Rough We know that cos 60Β° = 1/2 But we need (β1)/2 So, angle is in 2nd and 3rd quadrant ΞΈ = 60Β° 180 β ΞΈ = 180 β 60 = 120Β° = 120 Γ π/180 = 2π/3 x = 1/2 (2nΟ Β± 2π/3) x = nΟ Β± π/3 where n β Z Hence General Solution is For sin3x = 0, x = ππ /π OR For cos 2x = (β1)/2 , x = nΟ Β± π /π where n β Z