А) Решите cos2x+sin^2x=0,75 б) Найдите корни этого уравнения принадлежащие [пи;5пи/2]...

1 - 2Sin²x + Sin²x = 3/4
- Sin²x = 3/4 -1
Sin²x = 1/4
Sinx = +-1/2
a) Sinx = 1/2 б) Sin x = -2
x = (-1)^n π/6 + nπ, n ∈Z x = (-1)^(m+1) π/6 + mπ, m ∈ Z
[π; 5/2π] [π; 5π/2]
n = 2 m = 1, x = π/6 + π = 7π/6
x = π/6 +2π = m = 2 , x = -π/6 + 2π = 5π/6
=13π/6