sin2x(2cos2x-1)=cos2x(根号3-2sin2x) x大于等于0小于2π

1个回答

  • sin2x(2cos2x-1)=cos2x(√3-2sin2x)

    2sin2xcos2x-sin2x=√3cos2x-2sin2xcos2x

    4sin2xcos2x=√3cos2x+sin2x

    2sin4x=2(√3/2*cos2x+1/2*sin2x)

    sin4x=(sinπ/3*cos2x+cosπ/3*sin2x)

    sin4x=sin(π/3+2x)

    4x=2kπ+π/3+2x 或4x=2kπ+π-(π/3+2x)

    由4x=2kπ+π/3+2x得

    x=kπ+π/6

    k=0时,x=π/6;k=1时,x=7π/6;k=2时,x不在[0,2π)内,所以不考虑

    由4x=2kπ+π-(π/3+2x)得

    x=kπ/3+π/9

    k=0时,x=π/9;k=1时,x=4π/9;k=2时,x=7π/9;k=3时,x=10π/9;k=4时,x=14π/9;k=5时,x=17π/9,k=6时,x不在[0,2π)内,所以不考虑

    所以x值可为π/9,π/6,4π/9,7π/9,10π/9,7π/6,14π/9,17π/9