Question

在√3sinx+cosx=2a-3中，a的取值范围是多少

Answer 1

√3sinx+cosx=2a-3

(√3sinx+cosx)/2=(2a-3)/2

√3/2sinx+1/2cosx=(2a-3)/2
sinxcosπ/6+cosxsinπ/6=(2a-3)/2
sin(x+π/6)=(2a-3)/2
∵-1<=sin(x+π/6)<=1
∴-1<=(2a-3)/2<=1
-2<=2a-3<=2
1<=2a<=5
1/2<=a<=5/2

Answer 2

√3sinx+cosx=2(√3/2sinx+1/2cosx)=2(sinxcosπ/6+cosxsinπ/6)=2sin(x+π/6)
∴sin(x+π/6)=a-3/2
∵-1≤sin(x+π/6)≤1
∴-1≤a-3/2≤1
∴1/2≤a≤5/2