1）y=tanx-sinx
y'=sec²x-cosx=(1-cos³x)/cos²x
∵-1≤cosx≤1,cosx≠0
∴y'>0恒成立,即函数单调递增
令y=0,有1-cos³x=0,即1-cosx=0,cosx=1
∴极值点为x=2kπ,k∈Z
2）y=x+sinx
y'=1+cosx
∵-1≤cosx≤1
∴y'≥0恒成立,即函数单调递增
令y=0,有1+cosx=0,cosx=-1
∴极值点为x=π+2kπ,k∈Z
3）y=sinx+cosx=√2sin(x+π/4)
y'=√2cos(x+π/4)
当y'≥0时,有-π/2+2kπ≤x+π/4≤π/2+2kπ,即-3π/4+2kπ≤x≤π/4+2kπ
∴单调递增区间为x∈[-3π/4+2kπ,π/4+2kπ]
当y'≤0时,有π/2+2kπ≤x+π/4≤3π/2+2kπ,即π/4+2kπ≤x≤5π/4+2kπ
∴单调递减区间为x∈[π/4+2kπ,5π/4+2kπ]
令y=0,可得cos(x+π/4),即x+π/4=π/2+kπ,k∈Z
∴极值点为x=π/4+kπ,k∈Z