Question

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5cos²x+5cosx=1 - 3sin²x , здесь x∈[270°;450°]
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Answer 1

Решение:

5cos²x + 5cosx = 1 - 3sin²x        x∈[270°;450°]

5cos²x + 5cosx = 1-3(1-cos²x)

5cos²x + 5cosx - 1+3(1-cos²x) = 0

5cos²x + 5cosx - 1 + 3 - 3cos²x = 0

2cos²x + 5cos x + 2 = 0

Пусть cosx = a:

2a² + 5a + 2 = 0

D = 5² - 4×2×2 = 25 - 16 = 9

D>0, 2 корня

a₁ = -5 + √9/2×2 = -5+3/4 = -2/4 = -1/2

a₂ = -5 - √9/2×2 = -5-3/4 = -8/4 = -2  

cosx = -1/2                   или              cosx = -2

x = ±arccos(-1/2) + 2πn                    x∉R

x = ±(π - arccos 1/2) + 2πn

x = ±(π - π/3) + 2πn

x = ±2π/3 + 2πn

x = 2π/3 + 2πn, n∈Z

Пусть n = 0, n = 1, и n = 0,9  :

x = 2π/3 + 2π × 0 = 120° + 360°×0 = 120° + 0 = 120°

x = 2π/3 + 2π × 1 = 120° + 360°×1 = 120° + 360° = 480°

x = 2π/3 + 2π × 0,9 = 120° + 360°×0,9 = 120° + 324° = 444°

Ответ: x = 444°