Question

1) и 2) Найти y'

Answer 1

1) 2cos²x - sinx + 1 = 0
     2(1 - sin²x) - sinx + 1 = 0
      2 - 2sin²x - sinx + 1 = 0
      -2sin²x - sinx + 3 = 0
     2sin²x + sinx - 3 = 0
     y = sinx
    2y² + y - 3 = 0
    D= 1² - 4 · 2 · (-3) = 1 + 24 = 25; √25 = 5
    y1 = (-1 - 5)/(2 · 2) = -1,5
    y2 = (-1 + 5)/(2 · 2) = 1
    sinx = -1,5 - нет решений
    sinx = 1
    x = π/2 + 2πn, n ∈ Z 
2) y = -x⁵ + 2x³ - 3x² - 1
     y' = -5x⁴ + 6x² - 6x

Answer 2

Finding y'

To find the value of y', we need more context or information about the specific equation or problem related to y'. Could you please provide more details or clarify the specific equation or problem you are referring to? Once I have more information, I'll be able to assist you further.