5*(1/5)^cos2x=5^sin2x X=? X принадлежит (-7pi/2;2pi)

5 * (1/5)^cos2x = 5^sin2x, х ∈ (-7pi/2;2pi);5 * (5)^-cos2x = 5^sin2x;5^(-cos2x + 1) = 5^sin2x;-cos2x + 1 = sin2x;-(cos^2x - sin^2x) + 1 = 2sinxcosx;-cos^2x + sin^2x + 1 = 2sinxcosx;sin^2x + cos^2x - cos^2x + sin^2x = 2sinxcosx;2sin^2x = 2sinxcosx;2sin^2x - 2sinxcosx = 0;2sinx(sinx - cosx) = 0;2 sinx = 0 или sinx - cosx = 0;sinx = 0 или sinx = cosx;x = pin, n ∈ Z или tgx = 1;x = pi/4 + pin, n ∈ Z.-7pi/2 <= pin <= 2pi;-7/2 <= n <= 2;-3,5 <= n <= 2;n = -3, n = -2, n = -1, n = 0, n = 1, n = 2 => x = -3pi, x = -2pi, x = -pi, x = 0, x = pi, x = 2pi;-7pi/2 <= pi/4 + pin <= 2pi;-7pi/2 - pi/4 <= pin <= 2pi - pi/4;-14pi/4 - pi/4 <= pin <= 8pi/4 - pi/4;-13pi/4 <= pin <= 7pi/4;-13/4 <= n <= 7/4;n = -3, n = -2, n = -1, n = 0, n = 1 => x = pi/4 - 3pi = -11pi/4, x = pi/4 - 2pi = -7pi/4, x = pi/4 - pi = 3pi/4, x = pi/4, x = pi/4 + pi = 5pi/4.Ответ: -3pi, -2pi, -pi, 0, pi, 2pi или -11pi/4, -7pi/4, 3pi/4, pi/4, 5pi/4.