设a为常数,解方程cos(x-π/4)=sin(2x)+a
问题描述：
设a为常数,解方程cos(x-π/4)=sin(2x)+a
答
√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]
答
cos(x-π/4)=sin2x+acosxcos(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^...