Algebra 2 trigonometry help please?
回答 (8)
2018-04-13 2:37 pm
✔ 最佳答案
4 cos²θ - 1 = 2 cos²θ +12 cos²θ - 2 = 0
2 (cos²θ - 1) = 0
2 (cosθ - 1) (cosθ + 1) = 0
cosθ = 1 or cosθ = -1
θ = 0°, 180°, 360° for 0° ≤ θ ≤ 360°
General solution: θ = 2nπ for n = 0, 1, 2, 3 ….
2018-04-13 11:00 pm
4Cos^2(Th) - 2Cos^2(Th) = 1 + 1
2Cos^2(Th) = 2
Cos^2(th = 1
Cos(Th) = +/- 1
Th = Cos^-1(-1)
Th = 180 (pi)
&
Cos(Th) = 1
Th = Cos^-1(1)
Th = 0 & 360 (2pi)
2Cos^2(Th) = 2
Cos^2(th = 1
Cos(Th) = +/- 1
Th = Cos^-1(-1)
Th = 180 (pi)
&
Cos(Th) = 1
Th = Cos^-1(1)
Th = 0 & 360 (2pi)
2018-04-13 8:44 pm
4 cos^2 θ - 1 = 2 cos^2 θ + 1
4 cos^2 θ - 2 cos^2 θ - 2 =0
2 cos^2 θ - 2 = 0
2(cos^2 θ -1) = 0
cos^2 θ - 1 = 0
cos^2 θ = 1
cos θ = -1, 1
cos θ = -1
θ = pi + 2npi,
θ = (2n+1) pi, where n is any integer <---------
cos θ = 1
θ = 2pi n , where n is any integer <-----------
4 cos^2 θ - 2 cos^2 θ - 2 =0
2 cos^2 θ - 2 = 0
2(cos^2 θ -1) = 0
cos^2 θ - 1 = 0
cos^2 θ = 1
cos θ = -1, 1
cos θ = -1
θ = pi + 2npi,
θ = (2n+1) pi, where n is any integer <---------
cos θ = 1
θ = 2pi n , where n is any integer <-----------
2018-04-13 8:33 pm
4 cos²θ - 1= 2 cos²θ +1
i.e. 2 cos²θ - 2=0
or 2( cos²θ - 1)=0
i.e. 2( cosθ - 1)( cosθ + 1)=0
Hence either cosθ - 1=0 i.e. cosθ=1 hence θ = 2nπ for n= 0, 1, 2 , 3...........
or ( cosθ + 1) =0 cosθ= - 1 hence θ =π+ 2nπ = (2n+1) π for n= 0, 1, 2 , 3...........
Hence in general θ = nπ for n= 0, 1, 2 , 3...........
i.e. 2 cos²θ - 2=0
or 2( cos²θ - 1)=0
i.e. 2( cosθ - 1)( cosθ + 1)=0
Hence either cosθ - 1=0 i.e. cosθ=1 hence θ = 2nπ for n= 0, 1, 2 , 3...........
or ( cosθ + 1) =0 cosθ= - 1 hence θ =π+ 2nπ = (2n+1) π for n= 0, 1, 2 , 3...........
Hence in general θ = nπ for n= 0, 1, 2 , 3...........
2018-04-13 7:55 pm
4 cos^2(theta) - 1 = 2 cos^2(theta) + 1
2 cos^2(theta) - 2 = 0
2(cos^2(theta) - 1) = 0
-2 sin^2(theta) = 0 (or 2(cos(theta) + 1)(cos(theta) - 1) = 0)
sin(theta) = 0 (or cos(theta) = 1 or -1) --> theta = n * pi (assume that n is an integer).
2 cos^2(theta) - 2 = 0
2(cos^2(theta) - 1) = 0
-2 sin^2(theta) = 0 (or 2(cos(theta) + 1)(cos(theta) - 1) = 0)
sin(theta) = 0 (or cos(theta) = 1 or -1) --> theta = n * pi (assume that n is an integer).
2018-04-13 4:31 pm
2 cos ² Ө = 2
cos ² Ө = 1
cos Ө = ± 1
Ө = 0 ⁰ , 180 ⁰
cos ² Ө = 1
cos Ө = ± 1
Ө = 0 ⁰ , 180 ⁰
2018-04-13 3:03 pm
cos^2(θ) = 1
θ = πk where k ϵ Z
θ = πk where k ϵ Z
2018-04-13 2:47 pm
Cos what?
收錄日期: 2021-04-24 01:03:04
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