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標題:

sin tan cos

發問:

1.cos2x=0.4 2.√3 tan (x+30)=2 3.√2cos^2x-cosx=0 4.2sin^2-5sinx-3=0 5.√3tanx=2sinx 6.prove the identity tan (270+θ)=-1/tanθ

 

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最佳解答:

Suppose the range of x : 0 ≤ x ≤ 360 1. cos 2x = 0.4 2x = 66.4, 294, 426 or 654 x = 33.2, 147, 213 or 327 2. √3 tan(x + 30) = 2 tan (x + 30) = 2/√3 x + 30 = 49.1 or 229 x = 19.1 or 199 3. √2 cos^2 x - cos x = 0 cos x(√2 cos x - 1) = 0 cos x = 0 or cos x = 1/√2 x = 90, 270 or x = 45, 315 The solution for x are 45, 90, 270 or 315 4. 2 sin^2 x - 5 sin x - 3 =0 (2 sin x + 1)(sin x - 3) = 0 sin x = -1/2 or sin x = 3 (rej.) x = 210 or 330 5. √3 tan x = 2 sin x √3 sin x = 2 sin x cos x √3 sin x - 2 sin x cos x = 0 sin x(√3 - 2 cos x) = 0 sin x = 0 or √3 = 2 cos x x = 0,180 or cos x = √3 / 2 x = 0,180 or x = 30, 330 The solution for x are 0, 30, 180 or 330 6. L.H.S. = tan (270 + θ) = tan [90 + (180 + θ)] Let a = 180 + θ we have tan (90 + a) = -1/tan a = -1/tan (180 + θ) = -1/tan θ

其他解答:

1. cos 2x = 0.4 2x = 66.4 x=33.2 2. √3tan(x+30) = 2 tan(x+30) = 2/√3 x+30 = 49.1 x = 19.1 3.√2cos^2 x - cosx = 0 cosx(√2cosx-1) = 0 cosx = 0 or cosx = 1/√2 x = 90 or 45 5.√3tanx = 2sinx √3/cosx = 2 cosx = √3/2 x = 30 6.tan(270+θ) = -1/tanθ tan(270+θ) = -tan(90-θ) -tanθ = -tanθE2A5F59BAA12C031

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