f(cosx)=2x+1,f(sinx)=?

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f(cosx)=2x+1,f(sinx)=?f(cosx)=2x+1,f(sinx)=?f(cosx)=2x+1,f(sinx)=?f(sinx)=f[cos(π/2-x)]=2(π/2-x)+1=π

f(cosx)=2x+1,f(sinx)=?
f(cosx)=2x+1,f(sinx)=?

f(cosx)=2x+1,f(sinx)=?
f(sinx)
=f[cos(π/2-x)]
=2(π/2-x)+1
=π-2x+1

∫sin(x+π/3)dx
=-cos(x+π/3)
所以-cos(b+π/3)+cos(π/3+π/3)=0
cos(b+π/3)=cos2π/3
则b+π/3=2kπ±2π/3
b=2kπ+π/3,b=2kπ-π,且b≠-π/3