A)-π/3
B)-π/4
C)0
D)π/4
E)π/3
Which value of x in the interval -π/2 < x < π/2 satisfies the equation cos(x) = sec(x)?
2020-08-08 10:38 pm
回答 (3)
2020-08-08 10:49 pm
✔ 最佳答案
cos(x) = sec(x)cos(x) = 1/cos(x)
cos²(x) - 1 = 0
[cos(x) - 1] [cos(x) + 1] = 0
cos(x) = 1 or cos(x) = -1
x = 0 or x has no solution in -π/2 < x < π/2
The answer: C) 0
2020-08-09 12:54 am
secx = 1/cosx
so, cosx = 1/cosx
=> cos²x = 1
i.e. cosx = ±1
so, x = ±2nπ or ±π + 2nπ
With n = 0 we get x = 0 and ±π
Hence, x = 0 is the only solution for -π/2 < x < π/2
:)>
so, cosx = 1/cosx
=> cos²x = 1
i.e. cosx = ±1
so, x = ±2nπ or ±π + 2nπ
With n = 0 we get x = 0 and ±π
Hence, x = 0 is the only solution for -π/2 < x < π/2
:)>
2020-08-08 10:58 pm
Since sec = 1/cos, equation cos = sec is equivalent to equation cos = 1/cos, ie.,
cos^2 = 1. In (-pi/2,pi/2) this occurs only when x = 0. C) gives correct answer.
cos^2 = 1. In (-pi/2,pi/2) this occurs only when x = 0. C) gives correct answer.
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原文連結 [永久失效]:
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