Решить уравнение. / - черта дроби. 1/3х"в квадрате - 12 = 0

1/3х²-12=0

3х²=1/12

х²=(1/12 ): 3

х²=1/36

  х=±1/6

11/2*sin2x+1/2*sin12x-sin2x=0 1/2*sin12x-1/2*sin2x=0 1/2*(sin12x-sin2x)=0 1/2*sin5xcos7x=0 sin5x=0⇒5x=πn,n∈z⇒x=πn/5,n∈z cos7x=0⇒7x=π/2+πk,k∈z⇒x=π/14+πk/7,k∈z 2 3cos2x-1=cos2x/sin2x*2sin2x*cos2x 3cos2x-1=2cos²2x cos2x=a 2a²-3a+1=0 d=9-8=1 a1=(3-1)/4=1/2⇒cos2x=1/2⇒2x=+-π/3+2πn⇒x=+-π/6+πn,n∈z a2=(3+1)/4=1⇒cos2x=1⇒2x=2πk⇒x=πk,k∈z 3 2sin2x*cos2x=-2sin2x*sin3x 2sin2x*(cos2x+sin3x)=0 sin2x=0⇒2x=πn⇒x=πn/2,n∈z cos2x+sin3x=0 cos2x+cos(π/2-3x)=0 2cos(π/4-x/2)cos(5x/2-π/4)=0 2sin(x/2-π/4)cos(5x/2-π/4)=0 sin(x/2-π/4)=0⇒x/2-π/4=πn⇒x/2=π/4+πn⇒x=π/2+2πn,n∈z cos(5x/2-π/4)=0⇒5x/2-π/4=π/2+πk⇒5x/2=3π/4+πk⇒x=3π/10+2πk/5,k∈z 4 1/2*sin(-x)+1/2*sin7x+1/2*sin(-7x)+1/2*sin9x=0 1/2*sin9x-1/2*sinx=0 1/2*(sin9x-sinx)=0 1/2*2sin4xcos5x=0 sin4x=0⇒4x=πn⇒x=πn/4,n∈z cos5x=0⇒5x=π/2+πk⇒x=π/10+πk/5,k∈z

1)x²+3x=10 x(x+3)=10 x=10 или х+3=10                 х=10-3                 х=7 2)-1-3х=2х+1 5х=-2 х=2\5=0,4 3)-4\7х²+28=0 х²=-28\-(4\7) х²=49 х=7 4)х²+10=7х 7х-х²=10 х(7-х)=10 х=10 или 7-х=10                 х=-3 5)8х²-12х+4=0 х(8х-12)=-4 х=-4 или 8х-12=-4               8х=8               х=1