Find ALL exact radian solutions to the equation: sinx + 2sinxcosx = 0?
回答 (3)
2020-12-01 10:15 am
✔ 最佳答案
sinx + 2 sinx cosx = 0sinx (1 + 2 cosx) = 0
sinx = 0 or 1 + 2 cosx = 0
sinx = 0 or cosx = -1/2
x = nπ or x = 2nπ ± (2π/3) where n = 0, 1, 2 …..
For 0 ≤ x ≤ 2π
x = 0, 2π/3, π, 4π/3, 2π
2020-12-01 9:12 am
sin x + 2 sin x * cos x = 0
=> sin x ( 1 + 2 cos x ) = 0
=> sin x = 0 and cos x = - (1/2)
Mark that sin x = +ve and cos x = -ve, Hence x must lie in SECOND quadrant .
=> x = nπ or x = 2nπ + (2π/3) where n = 0, 1, 2 ….. ,
=> x = 0, (2π/3) , ....... Radians.
=> sin x ( 1 + 2 cos x ) = 0
=> sin x = 0 and cos x = - (1/2)
Mark that sin x = +ve and cos x = -ve, Hence x must lie in SECOND quadrant .
=> x = nπ or x = 2nπ + (2π/3) where n = 0, 1, 2 ….. ,
=> x = 0, (2π/3) , ....... Radians.
2020-12-01 9:05 am
Factor out sinx then solve for each factor.
收錄日期: 2021-04-23 23:06:42
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