解析：∵f(x)≤1+sinx
设h(x)=ax+cosx-sinx-1<=0 x∈[0,π]
h(x)=ax-√2sin(x-π/4)-1
令h’(x)=a-sinx-cosx=a-√2sin(x+π/4)=0
X1=√2/2a-π/4，x2=π-√2/2a-π/4
h’’(x)=-cosx-sinx=-√2sin(x+π/4)
当a>=√2π/4时
∴h(x)在x1处取极大值，在x2处取极小值
要满足f(x)≤1+sinx，只须满足h(x1)<=0
h(x1)=√2/2a^2-π/4a+√2cos(√2/2a)-1<=0
h(π)=aπ-√2sin(π-π/4)-1=aπ-2<=0==>a<=2/π
∵2/π<√2π/4，∴当a<=2/π时，x1<0
∴只要验证h(0)=-√2sin(-π/4)-1=0
当a<√2π/4时，x1<0(舍)，在x2处取极小值
∴h(0)=-√2sin(-π/4)-1=0
综上：要满足f(x)≤1+sinx，只须满足a<=2/π
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