Lời giải

`sqrt3*sin3x-2sinx*sin2x-cosx=0`

`<=>` `sqrt3*sin3x-2*(-1/2)*[cos(x+2x)-cos(2x-x)]-cosx=0`

`<=>` `sqrt3*sin3x+cos3x-cosx-cosx=0`

`<=>` `sqrt3*sin3x+cos3x=2cosx`

`<=>` `sqrt3/2*sin3x+1/2*cos3x=cosx`

`<=>` `cos3x*cos  pi/3+sin3x*sin  pi/3=cosx`

`<=>` `cos(3x-pi/3)=cosx`

`<=>` `[(3x-pi/3=x+k2pi \ (k in ZZ)),(3x-pi/3=-x+k2pi \ (k in ZZ)):}`

`<=>` `[(2x=pi/3+k2pi \ (k in ZZ)),(4x=pi/3+k2pi \ (k in ZZ)):}`

`<=>` `[(x = pi/6+kpi \ (k in ZZ)),(x=pi/12+(kpi)/2 \ (k in ZZ)):}`

Công thức

`***` `sina*sinb=-1/2*[cos(a+b)-cos(a-b)]`

`***` `cosa*cosb+sina*sinb=cos(a-b)`