1）f(π/4)=2cos(π/2)-cos(π/4)+2[sin(π/4)]^2
=0-√2/2+2(√2/2)^2
=1-(√2/2)
2）f(x)=2cos2x-cosx+2(sinx)^2
=2[2(cosx)^2-1]-cosx+2[1-(cosx)^2]
=4(cosx)^2-2-cosx+2-2(cosx)^2
=2(cosx)^2-cosx
=2(cosx-1/4)^2-1/8
所以,當(dāng)cosx=-1時(shí),f(x)取最大值,此時(shí)f(x)max=2+1=3
此時(shí),x=2kπ+π （k∈Z）