Question

设a为常数,解方程cox(x-45度）＝sin(2x)+a

Answer 1

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}