Ex 3.4, 9 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 17, 2020 by Teachoo
Last updated at Feb. 17, 2020 by Teachoo
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 ((๐ฅ + 5๐ฅ)/2) . cos ((๐ฅ โ 5๐ฅ)/2) + sin 3x = 0 2 sin (6๐ฅ/2) . cos ((โ4๐ฅ)/2) + sin 3x = 0 2 sin (3x) . cos (โ2x) + sin 3x = 0 We know that sin x + sin y = 2sin ((๐ฅ + ๐ฆ)/2) cos ((๐ฅ โ ๐ฆ)/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 = โ1 cos 2x = (โ1)/2 General solution is 3x = nฯ x = (๐๐ )/3 where n โ Z General solution for cos 2x = (โ๐)/๐ Let cos x = cos y cos 2x = cos 2y Given cos 2x = (โ1)/2 From (1) and (2) cos 2y = (โ1)/2 cos 2y = (โ1)/2 cos (2y) = cos (2๐/3) 2y = 2๐/3 General solution for cos 2x = cos 2y is 2x = 2nฯ ยฑ 2y Putting 2y = 2๐/3 2x = nฯ ยฑ 2๐/3 Rough We know that cos 60ยฐ = 1/2 But we need (โ1)/2 So, angle is in 2nd and 3rd quadrant ฮธ = 60ยฐ 180 โ ฮธ = 180 โ 60 = 120ยฐ = 120 ร ๐/180 = 2๐/3 x = 1/2 (2nฯ ยฑ 2๐/3) x = nฯ ยฑ ๐/3 where n โ Z Hence General Solution is For sin3x = 0, x = ๐๐ /๐ OR For cos 2x = (โ1)/2 , x = nฯ ยฑ ๐ /๐ where n โ Z
Ex 3.4
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Ex 3.4
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