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解三角方程2sin^2x+5sinxcosx+cos^2x=4
人气:283 ℃ 时间:2020-04-14 02:47:22
解答
2sin^2x+5sinxcosx+cos^2x=2sin^2x+5sinxcosx+(1-sin^2x)=sin^2x+5sinxcosx+1=-(1-2sin^2x-10sinxcosx-3)/2=-(cos2x-5sin2x-3)/2=4所以cos2x-5sin2x=-5-√26*sin[2x+arctan(-1/5)]=-52x+arctan(-1/5)=2kπ+arcsin(5/...
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