1.因?yàn)閇(sinwx)^2+(coswx)^2]^2=(sinwx)^4+(coswx)^4+2(sinwxcoswx)^2
=1
即f(x)=(sinwx)^4+(coswx)^4=1-2(sinwxcoswx)^2=1-[(sin2wx)^2]/2
=1-(1-cos4wx)/4
=(cos4wx)/4+3/4
又T/2=π/2,T=2π/(4w)=π
則w=1/2
2.由(1)則g(x)=(cos2x)/2+3/2+[sin(x+π/6)^2]
=(cos2x)/2+3/2+[1-cos(2x+π/3)]/2
=2+[(cos2x)-cos(2x+π/3)]/2
=2+cos(2x-π/3)/2
則g(x)max=5/2 此時(shí)cos(2x-π/3)=1
即2x-π/3=2kπ 即x=kπ+π/6 k屬于整數(shù)
同理g(x)min=3/2 此時(shí)cos(2x-π/3)=-1
即2x-π/3=2kπ +π 即x=kπ+2π/3 k屬于整數(shù)