Solve equation on the interval 0 to 2pi Sqrt2cos^2+cos=0?
回答 (2)
2017-10-17 9:50 pm
√2 cos²θ + cosθ = 0
cosθ (√2 cosθ + 1) = 0
cosθ = 0 or √2 cosθ + 1 = 0
cosθ = 0 or cosθ = -1/√2
θ = π/2, 3π/2 or θ = π - (π/4), π + (π/4)
θ = π/2, 3π/4, 5π/4, 3π/2
cosθ (√2 cosθ + 1) = 0
cosθ = 0 or √2 cosθ + 1 = 0
cosθ = 0 or cosθ = -1/√2
θ = π/2, 3π/2 or θ = π - (π/4), π + (π/4)
θ = π/2, 3π/4, 5π/4, 3π/2
2017-10-17 9:50 pm
sqrt(2) cos^2 x + cos x = 0
cos x( sqrt(2) cos x + 1) = 0
cos x = 0
x = pi/2, 3pi/2
sqrt(2) cos x + 1 = 0
sqrt(2) cos x = -1
cos x = -1/sqrt(2)
x = 3pi/4, 5pi/4
x= pi/2, 3pi/2, 3pi/4, 5pi/4
cos x( sqrt(2) cos x + 1) = 0
cos x = 0
x = pi/2, 3pi/2
sqrt(2) cos x + 1 = 0
sqrt(2) cos x = -1
cos x = -1/sqrt(2)
x = 3pi/4, 5pi/4
x= pi/2, 3pi/2, 3pi/4, 5pi/4
收錄日期: 2021-04-18 17:55:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171017134312AADMAQP