Sin(X/2) • cos(X/2) + 0,75 = 1


(8sin^4x - 6sin^2x+1)/(tg2x+корень 3) все это равно нулю

РешениеSin(X/2) • cos(X/2) + 0,75 = 1(1/2)* (2*Sin(X/2) • cos(X/2) =  1 - 3/4sinx = 1/2x = arcsin(1/2) + πk, k∈Zx = π/6 + πk, k∈Z(8sin^4x - 6sin^2x+1)/(tg2x+√3) = 08sin⁴x - 6sin²x + 1 = 0tg2x+√3 ≠ 08sin⁴x - 6sin²x + 1 = 0пусть sin²x = t8t² - 6t + 1 = 0D = 36 - 4*8*1 = 4t₁ = (6 - 2)/16 = 1/4t₂ = (6 + 2)/16 = 1/21) sin²x = 1/4sinx = - 1/2      x = (-1)^k*arcsin(-1/2) + πk, k∈Zx = (-1)^(k+1)*π/6 + πk, k∈Z или  sin x = 1/2x = (-1)^n*arcsin(1/2) + πn,n∈Zx = (-1)^n*π/6 + πk, k∈Z2) sin²x = 1/2sinx = - √2/2x = (-1)^m*arcsin(-√2/2) + πm, m∈Zx = (-1)^(m+1)*π/4 + πm, m∈Zsinx = √2/2x = (-1)^r*arcsin(√2/2) + πr, r∈Z x = (-1)^r*π/4 + πr, r∈Z