Question

设关于x的方程cos2x+根号3sin2x=k+1在 [0，π/2]内有两不同根m,n，求m+n的值及k的取值范围。

Answer 1

原式=2(1/2cos2x+根号3/2sin2x）=2sin(2x+π/6）=k+1sin(2x+π/6)=(k+1)/22x+π/6=arcsin(k+1)/2或π-arcsin(k+1)/2x=1/2arcsin(k+1)/2-π/12或x=5π/12-1/2arcsin(k+1)/2即m+n=π/3由题意知0<=（k+1）/2<1解得-1<=k<1