y=sin2x sin(2x π 3)

∵cos5π6=-cosπ6∴f(x)＝sin2xsinπ6−cos2xcos5π6=sin2xsinπ6+cos2xcosπ6=cos（2x−π6）令-π+2kπ≤2x−π6≤2kπ，得-5π12+

（1）∵函数f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ)(0<φ<π)，∴f（x）=12sin2xsin∅+1+cos2x2•cos∅-12cos∅=12s

设x-y=mx+y=n则6m-2n=143m+n=5解得m=2n=-1即x-y=2x+y=-1解得x=1/2y=-3/2

由(1)得3x+3y+2x-2y=365x+y=36(3)由(2)得4x+4y-3x+3y=-20x+7y=-20(4)(3)×7-(4)得34x=272∴x=8把x=8代入(3)得y=-4∴x=8y

{(x+y)/2+(x-y)/3=63(x+y)+2(x-y)=36(1)4(x+y)-3(x-y)=-20(2)由（1）*3+(2)*2得9(x+y)+6(x-y)+8(x+y)-6(x-y)=36

化简得：-x+7y=11①7x+3y=27②①式×7得：-7x+49y=77③②+③得：52y=104∴y=2代入①得：x=3∴x=3,y=2再问：亲，是代入法哦！再答：代入法①式得3x+3y-4x+

（I）f（x）=（2cos2x-1）cosφ+sin2xsinφ=cos2xcosφ+sin2xsinφ=cos（2x-φ），∵f（x）图象过点（π12，1）∴f（π12）=cos（π6-φ）=1，∵

X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1

因为(x-y)/(x+y)=3,则(x＋y)/(x－y)＝1/3则5(x-y)(x+y)-(x+y)/2(x-y)＝5*3-1/(3*2)=15-1/6=89/6

（1）f（x）=12sin2xsin∅+1+cos2x2cos∅− 12cos∅=12cos(2x−∅)（0＜∅＜π）∴2×π6−∅＝kπ∴∅＝π3（2）f（A）=12cos(2A−π3)＝

1、f(x)=1/2sin2xsinφ+(1+cos2x)/2cosφ-1/2cosφ=1/2sin2xsinφ+1/2cos2xcosφ=1/2cos(2x-φ)f(π/6)=1/2cos(π/3-

第二个方程是不是写错了2/（x+y）+3/（x-y）=6是这样吗

（1）原式f（x)=1/2sin2xsinφ+cos^xcosφ-1/2cosφ=1/2sin2xsinφ+cosφ（cos^x-1/2）=1/2sin2xsinφ+1/2cos2xcosφ=1/2c

@是α.. f(π/6)=√3/4sinα+3/4cosα-1/2cosα=√3/4sinα+1/4cosα=1/2(√3/2sinα+1/2cosα)=1/2sin(α+π/6)=1/2s

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

（Ⅰ）f（x）=12sin2xsinφ+cos2xcosφ-sin（π2+φ）=12sin2xsinφ+12（1+cos2x）cosφ-12cosφ=12sin2xsinφ+12cos2xcosφ=1

f(x)=1/2sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ),=1/2sin2xsinφ+1/2(1+cos2x)cosφ-1/2sin(π/2+φ),=1/2sin2xsin

∵cos5π6=-cosπ6∴f(x)＝sin2xsinπ6−cos2xcos5π6=sin2xsinπ6+cos2xcosπ6=cos（2x−π6）令-π+2kπ≤2x−π6≤2kπ，得-5π12+

很累-3x-2y