f(x)=sin2x+cos2x
=√2sin(2x+π/4)
所以T=2π/2=π
当2x+Pai/4=2kPai+Pai/2,即X=KPai+Pai/8时有最大值=√2
f(θ+π/8)=√2sin(2θ+π/4+π/4)
=√2cos2θ
=√2/3
cos2θ=1/3
θ锐角则sin2θ>0
sin²2θ+cos²2θ=1
sin2θ=2√2/3
tan2θ=2√2
tan2a=2tana/(1-tan^2a)=2根号2
tana=根号2-根号2tan^2a
根号2tan^2a+tana-根号2=0
tana=(-1+3)/2根号2=根号2/2