Ex 3.4, 6 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 12, 2020 by Teachoo
Last updated at Feb. 12, 2020 by Teachoo
Transcript
Ex 3.4, 6 Find the general solution of the equation cos 3x + cos x โ cos 2x = 0 cos 3x + cos x โ cos 2x = 0 (cos 3x + cos x) โ cos 2x = 0 2 cos ((3๐ฅ + ๐ฅ)/2) . cos ((3๐ฅ โ ๐ฅ)/2) โcos 2x = 0 2 cos (4๐ฅ/2) . cos (2๐ฅ/2) โcos 2x = 0 2 cos 2x . cos x โ cos 2x = 0 cos 2x (2cos x โ 1) = 0 We know that cos x + cos y = 2 cos ((๐ฅ + ๐ฆ)/2) cos ((๐ฅ โ ๐ฆ)/2) Replacing x by 3x & y by x Hence We find general solutions of both separately General solution for cos 2x = 0 Given cos 2x = 0 Thus, general solution is 2x = (2n + 1) ๐/2 x = (2n + 1) ๐ /๐ where n โ Z (2cos x โ 1) = 0 2cos x = 1 cos x = 1/2 General solution for cos x = ๐/๐ Let cos x = cos y Given cos x = 1/2 From (3) and (4) cos y = 1/2 cos y = cos (๐/3) y = ๐/3 Rough We know that cos 60ยฐ = 1/2 So, 60ยฐ = 60 ร ๐/180 = ๐/3 General solution for cos x = cos y is x = 2nฯ ยฑ y where n โ Z Putting y = ๐/3 x = 2nฯ ยฑ ๐/3 where n โ Z Hence General Solution is For cos 2x = 0, x = (2n + 1) ๐ /๐ Or For cos x = 1/2 , x = 2nฯ ยฑ ๐ /๐ where n โ Z
Ex 3.4
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