f(x)=msinx+√(2m-1)cosx
=√(m²+2m-1)sin(x+φ) tanφ=√(2m-1)/m
最小值为-√2=-√(m²+2m-1)
∴m²+2m-1=2→m=1 (m=-3,2m-1<0,舍去)
∴f(x)=sinx+cosx=√2sin(x+π/4)
x∈[-π,π/6]
x+π/4∈[-π+π/4,π/6+π/4]
[-π+π/4,-π/2)f(x)单调递减(-π/2,π/6+π/4)f(x)单调递增
最小值=√2sin(-π/2)=-√2
最大值=√2sin(π/6+π/4)=√2(sinπ/6cosπ/4+sinπ/4cosπ/6)
=√2(1/2·√2/2+√2/2·√3/2)=(1+√3)/2
∴f(x)∈[-√2,(1+√3)/2]
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