Calc 12!!?

👨🏻‍🏫 學術台數學

[匿名]

2020-04-20 12:21 pm

Solve, 0<theta<2pi,

cos(theta) = 0.825

回答 (3)

pingshek

2020-04-20 12:33 pm

✔ 最佳答案

cos(θ) > 0 when θ is in the first quadrant or the fourth quadrant.

cos(θ) = 0.825
θ = cos⁻¹(0.825)
θ = 0.60, 2π - 0.6
θ = 0.60, 5.68

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Jim

2020-04-20 12:53 pm

here's a graph, it helps to visualize

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[匿名]

2020-04-20 12:30 pm

arccos(0.825) = 0.6 rad
theta = 0.6, (2pi-0.6)
Just draw cos(x) and answers should makes sense.

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