数学例5（3）：已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) .若a∈[0,π]

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2sin(x-π/4)sin(x-π/4+π/2)
=1/2cos2x+√3/2sin2x+2sin(x-π/4)cos(x-π/4)
=1/2cos2x+√3sin2x+sin(2x-π/2)
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
f(a/2)=sin(2*a/2-π/6)=1/3
sin(a-π/6)=1/3
sinacosπ/6-cosasinπ/6=1/3
√3/2sina-1/2cosa=1/3
3√3sina-3cosa=2
cosa=√3sina-2/3
cos²a=3sin²a-4√3sina/3+4/9
∵cos²a=1-sin²a
∴3sin²a-4√3sina/3+4/9=1-sin²a
4sin²a-4√3sina/3-5/9=0
△=(-4√3/3)²-4*4*(-5/9)=16*8/9
sina=(4√3/3±√[16*8/9])/(2*4)
=(√3±2√2)/6
∵a∈[0,π]
∴sina=(√3+2√2)/6