1.Solve the trigonometric equation 2sin²θ+5sinθ-3=0 for 0<θ<360.
2.If θ is an acute angle and sinθ=cosθ,then cosθ=???
3.For 0<θ<360,how many roots does the equation 2cos²θ-5sinθ-4=0 have?
MATHS
2007-07-31 8:59 pm
回答 (3)
2007-07-31 9:12 pm
✔ 最佳答案
√1.2sin²θ + 5sinθ - 3 = 0
(2sinθ - 1)(sinθ + 3) = 0
sinθ = 1/2 or sinθ = -3 (rejected)
θ = 30 or 150
2.sinθ=cosθ
tanθ = 1 (cosθ =/= 0 for sinθ=cosθ)
θ = 45
cosθ = cos45 = 1/√2
3.2cos²θ-5sinθ-4=0
2 - 2sin²θ - 5sinθ - 4 = 0
2sin²θ + 5sinθ + 2 = 0
(2sinθ + 1)(sinθ + 2) = 0
sinθ = -1/2 or sinθ = -2(rejected)
Thus we have two roots for the equation. (210 and 330)
2007-07-31 9:13 pm
2sin θ - √3 = 0
2sinθ=√3
sinθ=√3/2
θ=60,120
2sinθ=√3
sinθ=√3/2
θ=60,120
2007-07-31 9:05 pm
1)
2sin θ - √3 = 0
2sinθ=√3
sinθ=√3/2
θ=60,120
I CAN'T help U beacuse i study chines maths.....i can't understand english maths
2sin θ - √3 = 0
2sinθ=√3
sinθ=√3/2
θ=60,120
I CAN'T help U beacuse i study chines maths.....i can't understand english maths
收錄日期: 2021-04-13 00:55:32
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