原帖由 kay199143 於 2009-5-5 10:50 PM 發表
cos3x=sinx求通解~~ sinx=cos3x=cos(±3x) = sin(pi/2 ± 3x+2npi)
x = pi/2 ± 3x+2npi, where n can be any integer.
Solve for x,
x = -pi/4 - npi
In [0, 2pi), x = 3pi/4, 7pi/4
or
x = pi/8 + npi/2
In [0, 2pi), x = pi/8, 5pi/8, 9pi/8, 13pi/8 |