解三角方程2sin^2x+5sinxcosx+cos^2x=4

题目

解三角方程2sin^2x+5sinxcosx+cos^2x=4

答案

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/...