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解下列方程 cos2x=cosx+sinx sin^4x-cos^4x=cosx+sinx
解下列方程 cos2x=cosx+sinx
sin^4x-cos^4x=cosx+sinx
人气:100 ℃ 时间:2019-10-08 12:37:00
解答
1)cos2x=cosx+sinx --->cos^2 x-sin^2 x=cosx+sinx---> (cosx+sinx)(cosx-sinx-1)=0---> cosx+sinx=0---> tgx=-1--> x= kπ-π/4,这里k为任意整数.orcosx-sinx-1=0--> cos(x+π/4)=1/√2--> x+π/4=2kπ+/-π/...
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