解三角方程2sin^2x+5sinxcosx+cos^2x=4
问题描述：
解三角方程2sin^2x+5sinxcosx+cos^2x=4
答
用降次把一三项平方去掉变成二倍角2x,中间项也变2x,最后利用asinx+bcosx=__合并便可解
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/...