Найти нули функций y=-x/8 y=x-5/4

Question

Алгебра: Найти нули функций y=-x/8 y=x-5/4

gabdrahmanova951 · Accepted Answer

2sin^2x - 3cosx - 3 = 0, х ∈ [pi; 3pi]; 2(1 - cos^2x) - 3cosx - 3 = 0; 2 - 2cos^2x - 3cosx - 3 = 0; -2cos^2x - 3cosx - 1 = 0; 2cos^2x + 3cosx + 1 = 0; Пусть cosx = t, тогда 2t^2 + 3t + 1 = 0; D = 9 - 4 * 2 * 1 = 1; t = (-3 +- 1)/ (2 * 2); t1 = -1/2, t2 = -1; cosx = -1/2, x = +- arccos(-1/2) + 2pi * n, n ∈ N, x = +- 2pi/3 + 2pi * n, n ∈ N; cosx = -1, x = pi + 2pin, n ∈ N; pi = pi + 2pin = 3pi; 0 = 2pin = 2pi; 0 = n = 1; n = 1 = x = pi + 2pi = 3pi; n = 0 = x = pi; pi = - 2pi/3 + 2pi * n = 3pi; pi...

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