Решить уравнение: 2cos^2x+sinx+1=0

2Cos²x + Sinx + 1 = 0

2 (1 - Sin²x) + Sinx + 1 = 0

2 - 2Sin²x + Sinx + 1 = 0

2Sin²x - Sinx - 3 = 0

Sinx = m, - 1 ≤ x ≤1

2m² - m - 3 = 0

D = (-1) ² - 4 * 2 * (- 3) = 1 + 24 = 25

m₁ = (1 + √25) / 4 = (1 + 5) / 4 = 1,5 - не подходит

m₂ = (1 - √25) / 4 = (1 - 5) / 4 = - 1

Sinx = - 1

x = - π/2 + 2πn, n ∈ z