设a为常数,解方程cos(x-π/4)=sin(2x)+a
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设a为常数,解方程cos(x-π/4)=sin(2x)+a
设a为常数,解方程cos(x-π/4)=sin(2x)+a
设a为常数,解方程cos(x-π/4)=sin(2x)+a
cos(x-π/4)=sin2x+a
cosxcos(π/4)+sinxsin(π/4)=2sinxcosx+a
(根号2/2)*cosx+(根号2/2)*sinx=2sinxcosx+a
(根号2/2)(cosx+sinx)=2sinxcosx+a
两边同时平方,得:
(1/2)(cosx+sinx)^2=(2sinxcosx+a)^2
(1/2)[sin^2(x)+cos^2(x)+2sinxcosx]=[4sin^2(x)cos^2(x)+a^2+4asinxcosx]
由于sin^2(x)+cos^2(x)=1
则:
(1/2)*[1+2sinxcosx]=[4sin^2(x)cos^2(x)+a^2+4asinxcosx]
设T=sinxcosx
=(2sinxcosx)/2
=(1/2)sin2x
则:
(1+2T)/2=(4T^2+a^2+4aT)
8T^2+(8a-2)T+(2a^2-1)=0
则由求根公式,得:
T=(1/2)sin2x
=[(1-4a)+根号(9-8a)]/16
或 =[(1-4a)-根号(9-8a)]/16
则:sin2x=[(1-4a)+根号(9-8a)]/8
或 =[(1-4a)-根号(9-8a)]/8
则x=arcsin{[(1-4a)+根号(9-8a)]/4}
或=arcsin{[(1-4a)-根号(9-8a)]/4}
√2(cosx+sinx)/2=2sinxcosx+a
cosx+sinx=2√2sinxcosx+√2 a
两边平方：
1+2sinxcosx=8(sinxcosx)^2+8asinxcosx+2a^2
设sinxcosx=t
1+2t=8t^2+8at+2a^2
8t^2+(8a-2)t+2a^2-1=0
解得t=(1-4a±2√(9-8a))/8=sin(2x)/2
故x=arcsin[(1-4a±2√(9-8a))/4]