In the interval 0°≤θ <360°, solve the equation 5 cos θ = 2 sec θ - 3 algebraically for all values of θ, to the nearest tenth of a degree.?
回答 (3)
2016-06-16 7:00 pm
5 cosθ = 2 secθ - 3
5 cosθ = 2 × (1/cosθ) - 3
5 cosθ × cosθ = [2 × (1/cosθ) - 3] × cosθ
5 cos²θ = 2 - 3 cosθ
5 cos²θ +3 cosθ - 2 = 0
(5 cosθ - 2)(cosθ + 1) = 0
5 cosθ - 2 = 0 or cosθ = -1
cosθ = 2/5 or cosθ = -1
θ = 66.4°, (360 - 66.4°) or θ = 180°
θ = 66.4°, 180°, 293.6°
5 cosθ = 2 × (1/cosθ) - 3
5 cosθ × cosθ = [2 × (1/cosθ) - 3] × cosθ
5 cos²θ = 2 - 3 cosθ
5 cos²θ +3 cosθ - 2 = 0
(5 cosθ - 2)(cosθ + 1) = 0
5 cosθ - 2 = 0 or cosθ = -1
cosθ = 2/5 or cosθ = -1
θ = 66.4°, (360 - 66.4°) or θ = 180°
θ = 66.4°, 180°, 293.6°
2016-06-16 6:49 pm
5cos(t) = 2/cos(t) - 3
5cos(t)^2 = 2 - 3cos(t)
5cos(t)^2 + 3cos(t) - 2 = 0
cos(t) = (-3 +/- sqrt(9 + 40)) / 10
cos(t) = (-3 +/- 7) / 10
cos(t) = -10/10 , 4/10
cos(t) = -1 , 2/5
cos(t) = 180 , arccos(2/5) , 360 - arccos(2/5)
5cos(t)^2 = 2 - 3cos(t)
5cos(t)^2 + 3cos(t) - 2 = 0
cos(t) = (-3 +/- sqrt(9 + 40)) / 10
cos(t) = (-3 +/- 7) / 10
cos(t) = -10/10 , 4/10
cos(t) = -1 , 2/5
cos(t) = 180 , arccos(2/5) , 360 - arccos(2/5)
2016-06-16 6:43 pm
Yes it does.
收錄日期: 2021-04-18 15:13:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160616104116AAqtG0B