решить уравнение 
tgx-4sin2x-2sin^2x=2cos^2x-ctgx

tg(x) = sinx/cosxctg(x) = cosx/sinxsin(2x) = 2sinx*cosxsin^2(x) + cos^2(x) = 1(sinx/cosx) - 8sinx*cosx = 2(sin^2(x) + cos^2(x)) - (cosx/sinx)(sinx/cosx) - 8sinx*cosx + (cosx/sinx) - 2 = 0(sin^2(x) + cos^2(x))/(sinx*cosx) - 8sinx*cosx - 2 = 0(1 - 8(sinx*cosx)^2 - 2sinx*cosx)/(sinx*cosx) = 01 - 2*(2sinx*cosx)^2 - sin(2x) = 01 - 2sin^2(2x) - sin(2x) = 0Замена: sin(2x) = t, t∈[-1;1]-2t^2 - t + 1 = 02t^2 + t - 1 = 0, D = 1 + 4*2 = 9t1 = (-1 + 3)/4 = 2/4 = 0.5t2 = (-1 - 3)/4 = -4/4 = -11) sin(2x) = 0.52x = π/6 + 2πk, x=π/12 + πk2x = 5π/6 + 2πk, x=5π/12 + πk2) sin(2x) = -12x = π/2 + 2πk, x=π/4 + πk