Find all solutions of the equation in the interval (0, 2π)
cos^2(x)=2-2sin(x)
Write your answer in radians in terms of π
If there is more than one solution, separate them with commas.
Please help
Find all solutions of the equation in the interval (0, 2π)?
回答 (3)
2015-10-12 10:21 pm
✔ 最佳答案
cos^2(x) = 1 - sin^2(x)1 - sin^2(x) = 2 - 2 sin x
sin^2(x) - 2sin(x) + 1 = 0
(sin(x) - 1)^2 = 0
sin(x) - 1 = 0
sin(x) = 1
x = pi/2
2015-10-12 10:20 pm
1-sin^2(x) = 2-2sin(x)
sin^2(x)-2sin(x)+1 = 0
(sinx-1)^2 = 0
sinx = 1
x = pi/2
sin^2(x)-2sin(x)+1 = 0
(sinx-1)^2 = 0
sinx = 1
x = pi/2
2015-10-12 10:20 pm
cos(x)^2 = 2 - 2sin(x)
1 - sin(x)^2 = 2 - 2sin(x)
sin(x)^2 - 2sin(x) + 2 - 1 = 0
sin(x)^2 - 2sin(x) + 1 = 0
(sin(x) - 1)^2 = 0
sin(x) - 1 = 0
sin(x) = 1
x = pi/2
1 - sin(x)^2 = 2 - 2sin(x)
sin(x)^2 - 2sin(x) + 2 - 1 = 0
sin(x)^2 - 2sin(x) + 1 = 0
(sin(x) - 1)^2 = 0
sin(x) - 1 = 0
sin(x) = 1
x = pi/2
收錄日期: 2021-04-21 14:36:30
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