Question

$4cos^{2} x+4sinx-1$

Answer 1

4cos²x + 4sinx - 1 = 0
4(1 - sin²x) + 4sinx - 1 = 0
4 - 4sin²x + 4sinx - 1 = 0
-4sin²x + 4sinx + 3 = 0
4sin²x - 4sinx - 3 = 0
t = sinx
4t² - 4t - 3 = 0
D = (-4)² - 4 · 4 · (-3) = 16 + 48 = 64; √64 = 8
t₁ = (4 - 8)/(2 · 4) = -4/8 = -1/2
t₂ = (4 + 8)/(2 · 4)  = 12/8 = 4/3
sinx = 4/3 -  нет решений, т.к. 4/3 > 1, a |sinx| ≤ 1
sinx = -1/2
x = (-1)ⁿ · (-π/6) +πn, n ∈ Z
x = (-1)ⁿ⁺¹ · π/6 +πn, n ∈ Z