Solve for θ: 2 cos 3θ = -1, interval 0 to 2π Please help :)?
回答 (4)
2016-02-21 2:55 pm
✔ 最佳答案
2 cos 3θ = -1cos 3θ = -1/2
Since it is 3θ (and not θ) , we need to go around the circle three times
3θ = 2pi/3 , 4pi/3 , 2pi/3+2pi, 4pi/3+2pi , 2pi/3 + 4pi , 4pi/3+4pi
3θ = 2pi/3, 4pi/3 ,8pi/3, 10pi/3, 14pi/3, 16pi/3
θ = 2pi/9, 4pi/9 ,8pi/9, 10pi/9, 14pi/9, 16pi/9
2016-02-21 2:51 pm
0 ≤ θ ≤ 2π
0 ≤ 3θ ≤ 6π
2 cos 3θ = -1
cos 3θ = -1/2
3θ = π ± (π/3), 2π + π ± (π/3), 4π + π ± (π/3)
3θ = 2π/3, 4π/3, 8π/3, 10π/3, 14π/3, 16π/3
θ = 2π/9, 4π/9, 8π/9, 10π/9, 14π/9, 16π/9
0 ≤ 3θ ≤ 6π
2 cos 3θ = -1
cos 3θ = -1/2
3θ = π ± (π/3), 2π + π ± (π/3), 4π + π ± (π/3)
3θ = 2π/3, 4π/3, 8π/3, 10π/3, 14π/3, 16π/3
θ = 2π/9, 4π/9, 8π/9, 10π/9, 14π/9, 16π/9
2016-02-21 2:43 pm
Assuming cos 3θ means cos(3θ), not cos3 * θ or cos^3θ,
2cos(3θ) = -1
cos(3θ) = -1/2
3θ = 2pi/3 + 2npi or 3θ = 4pi/3 + 2npi for all integers, n
θ = 2pi/9 + 2npi/3 or θ = 4pi/9 + 2npi/3 for all integers, n
We want θ in [0, 2pi) so
θ = 2pi/9 or θ = 8pi/9 or θ = 14pi/9 or θ = 4pi/9 or θ = 10pi/9 or θ = 16pi/9
2cos(3θ) = -1
cos(3θ) = -1/2
3θ = 2pi/3 + 2npi or 3θ = 4pi/3 + 2npi for all integers, n
θ = 2pi/9 + 2npi/3 or θ = 4pi/9 + 2npi/3 for all integers, n
We want θ in [0, 2pi) so
θ = 2pi/9 or θ = 8pi/9 or θ = 14pi/9 or θ = 4pi/9 or θ = 10pi/9 or θ = 16pi/9
2016-02-21 3:23 pm
cos 3Ө = - 1
3Ө = π , 3π , 5π
Ө = π/3 , π , 5π/3
3Ө = π , 3π , 5π
Ө = π/3 , π , 5π/3
收錄日期: 2021-04-18 14:28:17
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