由題意得f(x)=f(x+π),且f(x)=f(-x)
則f(-x)=f(-x+π)=f(x)
f(x)以π/2為對(duì)稱軸
由x∈[0,π/2]時(shí),f（x）=sinx
則當(dāng)x∈[π/2,π]時(shí),-x+π∈[0,π/2],f(x)=f(π-x)=sin(π-x)=sinx
故當(dāng)x∈[0,π]時(shí),f（x）=sinx
(1) 當(dāng)x∈[-π,0]時(shí),-x∈[0,π],f（-x）=-sinx=f(x)
所以 當(dāng)x∈[-π,0]時(shí),f（x）=-sinx.
(2) 由(1)可以知道,對(duì)x∈R,f（x）=|sinx|
當(dāng)f（x）≥1/2時(shí),x∈[π/6+kπ,5π/6+kπ],k∈Z