由辅助角公式asinα+bcosα=√(a²+b²)sin(α+φ)asin(x+π/3)+(a-1)cos(x+π/3)=√[a²+(a-1)²]sin(x+π/3+φ)=2a-1所以sin(x+π/3+φ)=(2a-1)/√[a²+(a-1)²]由|sin(x+π/3+φ)|≤1得|(2a-1)/√[a²+(a-1)²]|≤1解得0≤a≤1。
