解方程:(sin2x+cos2x)/(1-sin^2x-2cos2x)=2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/08 04:34:28
解方程:(sin2x+cos2x)/(1-sin^2x-2cos2x)=2
X属于(π/2,π)
X属于(π/2,π)
sin2x+cos2x
=1-2sin^2x+2sinxcosx
2-2sin^2x-4cos2x
=-2sin^2x+2-4+8sin^2x
=-2+6sin^2x
即,8sin^2x-2sinxcosx-3=0
5sin^2x-2sinxcosx-3cos^2x=0
(5sinx+3cosx)(sinx-cosx)=0
因为,X属于(π/2,π)
所以5sinx=-3cosx
tanx=-3/5
x=π-arctan(3/5)
=1-2sin^2x+2sinxcosx
2-2sin^2x-4cos2x
=-2sin^2x+2-4+8sin^2x
=-2+6sin^2x
即,8sin^2x-2sinxcosx-3=0
5sin^2x-2sinxcosx-3cos^2x=0
(5sinx+3cosx)(sinx-cosx)=0
因为,X属于(π/2,π)
所以5sinx=-3cosx
tanx=-3/5
x=π-arctan(3/5)
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