solve (in radians) for all values of x between 0 and 2pi rad.. sin(2x)cosx-cos(2x)sinx=0?
回答 (4)
2016-11-29 5:47 pm
sin(2x) cosx − cos(2x) sinx = 0
(2 sinx cosx) cosx − (2 cos²x − 1) sinx = 0
2 sinx cos²x − 2 sinx cos²x + sinx = 0
sinx = 0
x = 0, π, 2π
2016-11-29 6:54 pm
2 sin x cos²x - [ 2 cos²x - 1 ] sin x = 0
sin x = 0
x = 0 , π
sin x = 0
x = 0 , π
2016-11-29 5:47 pm
sin(2x)cos(x) - cos(2x)sin(x) = 0
[2sin(x)cos(x)]cos(x) - [2cos²(x) - 1]sin(x) = 0
sin(x)[2cos²(x) - 2cos²(x) + 1] = 0
sin(x) = 0
x = 0, π, or 2π
[2sin(x)cos(x)]cos(x) - [2cos²(x) - 1]sin(x) = 0
sin(x)[2cos²(x) - 2cos²(x) + 1] = 0
sin(x) = 0
x = 0, π, or 2π
2016-11-29 5:46 pm
use sin(a-b) = sin a cos b - sin b cos a
then
sin(2x)cosx-cos(2x)sinx=0
sin(2x-x) = 0
sin(x) = 0 => x = n*pi, where n is integer.
then
sin(2x)cosx-cos(2x)sinx=0
sin(2x-x) = 0
sin(x) = 0 => x = n*pi, where n is integer.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161129094215AAEmymo