求二元函数方程组sin(x-y)=0;cos(x+y)=0的解
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求二元函数方程组sin(x-y)=0;cos(x+y)=0的解
sin(x-y)=0
sinxcosy- cosxsiny =0
cos(x+y)=0
cosxcosy-sinxsiny = 0
sinxcosy- cosxsiny = cosxcosy-sinxsiny
sinx-cosxtany= cosx- sinxtany
sinx-cosx= -tany(sinx-cosx)
(sinx-cosx)(1-tany)=0
sinx = cos x
x = 2kπ+ π/4
or tany = 1
or y = kπ + π/4
k=0,(+/-)1,(+/-)2,.
sinxcosy- cosxsiny =0
cos(x+y)=0
cosxcosy-sinxsiny = 0
sinxcosy- cosxsiny = cosxcosy-sinxsiny
sinx-cosxtany= cosx- sinxtany
sinx-cosx= -tany(sinx-cosx)
(sinx-cosx)(1-tany)=0
sinx = cos x
x = 2kπ+ π/4
or tany = 1
or y = kπ + π/4
k=0,(+/-)1,(+/-)2,.
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