sin3x-√3cos2x-sinx=0

Sin3x-√3cos2x-sinx=0sin2x cosx+sinx cos2x-√3cos2x-sinx=02sinx cos²x+sinx( cos2x -1)-√3(cos²x-sin²x)=02sinx (1-sin²x)-sinx( 2sin²x)-√3(1-2sin²x)=0-4sin³x+2sinx+2√3sin²x-√3=0sinx=t      It I≤1-4t³+2√3 t²+2t-√3=0(4t³-2√3 t²)-(2t-√3)=02t²(2t-√3)-(2t-√3)=0     (2t-√3)(2t²-1)=0 ⇒t1=√3/2    t2=1/√2  t3= -1/√2t1=√3/2sinx=√3/2  ⇔  x=(-1)^n ·π/3 +πn, n∈Z t2=1/√2  t3= -1/√2    sin²x =1/2  ⇔2sin²x=1  1-cos2x=1 ⇔cos2x=02x=π/2+πn, n∈Z  x=π/4+πn/2, n∈Z