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Диана201198

16 июля 2013 г., 8:07:14 (10 лет назад)

1) cos2x -1 = sinx

cos²x - sin²x - sinx - 1 = 0

1-sin²x - sin²x - sinx - 1 = 0

-2sin²x - sinx = 0

2sin²x + sinx = 0

sinx(2sinx+1) = 0

1) sinx = 0

x = πn

2) 2sinx + 1 = 0

sinx = -1/2

x = (-1)^n+1 *π/6 + πn

2) sinx +√3cosx = 0

делим обе части на 2

sinx*1/2 + √3cosx*1/2 = 0

sin(x+π/3) = 0

(x+π/3) = πn

x = -π/3 + πn

3) cos2x = cosx-1

cos²x - sin²x - cosx + 1 = 0

cos²x -(1-cos²x) - cosx + 1 = 0

2cos²x  - cosx = 0

cosx(2cosx - 1) = 0

1) cosx = 0

x = π/2 + πn

2) 2cosx - 1 = 0

cosx = 1/2

x = ±π/3 + 2πn