Question

若方程sinx+√3 cosx+a=0在x∈[0,π]内有解,则实数a的取值范围

Answer 1

∵sinx+√3 cosx=-a ∴ 2(sinx/2+√3 cosx/2)=-a

∴2(sinxcosπ/3+cosxsinπ/3)=-a ∴sin(x+π/3)=-a/2

∵x∈[0,π] ∴x+π/3∈[π/3,4π/3] ∴-√3/2＜-a/2＜1 ∴-2＜a＜√3