Sin(x/2) • cos(x/2) + 0,75 = 1 (8sin^4x - 6sin^2x+1)/(tg2x+корень 3) все это равно нулю

Решение
Sin(X/2) • cos(X/2) + 0,75 = 1
(1/2)* (2*Sin(X/2) • cos(X/2) =  1 - 3/4
sinx = 1/2
x = arcsin(1/2) + πk, k∈Z
x = π/6 + πk, k∈Z

(8sin^4x - 6sin^2x+1)/(tg2x+√3) = 0

8sin⁴x - 6sin²x + 1 = 0
tg2x+√3 ≠ 0

8sin⁴x - 6sin²x + 1 = 0
пусть sin²x = t
8t² - 6t + 1 = 0
D = 36 - 4*8*1 = 4
t₁ = (6 - 2)/16 = 1/4
t₂ = (6 + 2)/16 = 1/2
1) sin²x = 1/4
sinx = - 1/2      
x = (-1)^k*arcsin(-1/2) + πk, k∈Z
x = (-1)^(k+1)*π/6 + πk, k∈Z
 или  sin x = 1/2
x = (-1)^n*arcsin(1/2) + πn,n∈Z
x = (-1)^n*π/6 + πk, k∈Z
2) sin²x = 1/2
sinx = - √2/2
x = (-1)^m*arcsin(-√2/2) + πm, m∈Z
x = (-1)^(m+1)*π/4 + πm, m∈Z
sinx = √2/2
x = (-1)^r*arcsin(√2/2) + πr, r∈Z 
x = (-1)^r*π/4 + πr, r∈Z