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解方程 2sin^2x-3sinxcosx-2cos^2x=0
人气:176 ℃ 时间:2020-03-07 01:41:57
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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