(1)f(x)=msinx+√2cosx =√(m²+2){[m/√(m²+2)]sinx+[√2/√(m²+2)]cosx} =√(m²+2)sin(x+α)
其中cosα=m/√(m²+2),
sinα=√2/√(m²+2)
可见f(x)的最大值为√(m²+2)=2,
解得m=√2
所以原函数为 f(x)=√2sinx+√2cosx=2sin(x+π/4) 其单调减区间满足 π/2+2kπ≤x+π/4≤3π/2+2kπ,k∈Z
即减区间为 2kπ+π/4≤x≤2kπ+5π/4
(2)f(B)=√3,
则2sin(B+π/4)=√3
所以B+π/4=π/3
解得 B=π/3-π/4=π/12