Ex 3.4, 6 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Feb. 12, 2020 by
Last updated at Feb. 12, 2020 by
Transcript
Ex 3.4, 6 Find the general solution of the equation cos 3x + cos x β cos 2x = 0 cos 3x + cos x β cos 2x = 0 (cos 3x + cos x) β cos 2x = 0 2 cos ((3π₯ + π₯)/2) . cos ((3π₯ β π₯)/2) βcos 2x = 0 2 cos (4π₯/2) . cos (2π₯/2) βcos 2x = 0 2 cos 2x . cos x β cos 2x = 0 cos 2x (2cos x β 1) = 0 We know that cos x + cos y = 2 cos ((π₯ + π¦)/2) cos ((π₯ β π¦)/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) π/2 x = (2n + 1) π /π where n β Z (2cos x β 1) = 0 2cos x = 1 cos x = 1/2 General solution for cos x = π/π Let cos x = cos y Given cos x = 1/2 From (3) and (4) cos y = 1/2 cos y = cos (π/3) y = π/3 Rough We know that cos 60Β° = 1/2 So, 60Β° = 60 Γ π/180 = π/3 General solution for cos x = cos y is x = 2nΟ Β± y where n β Z Putting y = π/3 x = 2nΟ Β± π/3 where n β Z Hence General Solution is For cos 2x = 0, x = (2n + 1) π /π Or For cos x = 1/2 , x = 2nΟ Β± π /π where n β Z
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