解方程 2sin^2x-3sinxcosx-2cos^2x=0
来源:学生作业帮 编辑:神马作文网作业帮 分类:综合作业 时间:2024/09/20 22:41:50
解方程 2sin^2x-3sinxcosx-2cos^2x=0
2sin^2x+5sinxcosx+cos^2x
=2sin^2x+5sinxcosx+(1-sin^2x)
=sin^2x+5sinxcosx+1
=-(1-2sin^2x-10sinxcosx-3)/2
=-(cos2x-5sin2x-3)/2=4
所以cos2x-5sin2x=-5
-√26*sin[2x+arctan(-1/5)]=-5
2x+arctan(-1/5)=2kπ+arcsin(5/√26)
或2x+arctan(-1/5)=2kπ+π-arcsin(5/√26)
所以x=[2kπ+arcsin(5/√26)+arctan(1/5)]/2
或x=[2kπ+π-arcsin(5/√26)+arctan(1/5)]/2
又arcsin(5/√26)+arctan(1/5)=π/2
-arcsin(5/√26)+arctan(1/5)=-arctan(12/5)
所以x=kπ+π/4
或x=[2kπ+π-arctan(12/5)]/2
=2sin^2x+5sinxcosx+(1-sin^2x)
=sin^2x+5sinxcosx+1
=-(1-2sin^2x-10sinxcosx-3)/2
=-(cos2x-5sin2x-3)/2=4
所以cos2x-5sin2x=-5
-√26*sin[2x+arctan(-1/5)]=-5
2x+arctan(-1/5)=2kπ+arcsin(5/√26)
或2x+arctan(-1/5)=2kπ+π-arcsin(5/√26)
所以x=[2kπ+arcsin(5/√26)+arctan(1/5)]/2
或x=[2kπ+π-arcsin(5/√26)+arctan(1/5)]/2
又arcsin(5/√26)+arctan(1/5)=π/2
-arcsin(5/√26)+arctan(1/5)=-arctan(12/5)
所以x=kπ+π/4
或x=[2kπ+π-arctan(12/5)]/2
解方程:sin^2x+2sinxcosx-3cos^3x=-2
方程sin^2x+2sinxcosx-2cos^2x-m=0恒有解
sin²X-sinxcosx+2cos²x=
sin²x+sinxcosx+2cos²x=
若关于x的方程sin^x+2sinxcosx-2cos^x=a有实数解.
秒回答秒采纳 sin^2x-2sinxcosx-3cos^2x=0
已知函数f(x)=sin^x+根号3sinxcosx+2cos^x,x属于R
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