Sin3x-√3cos2x-sinx=0

Sin3x-√3cos2x-sinx=0

sin2x cosx + sinx cos2x-√3cos2x-sinx=0

2sinx cos²x+sinx (cos2x - 1) - √3 (cos²x-sin²x) = 0

2sinx (1-sin²x) - sinx (2sin²x) - √3 (1-2sin²x) = 0

-4sin³x+2sinx+2√3sin²x-√3=0

sinx=t It I≤1

-4t³+2√3 t²+2t-√3=0

(4t³-2√3 t²) - (2t-√3) = 0

2t² (2t-√3) - (2t-√3) = 0 (2t-√3) (2t²-1) = 0 ⇒t1=√3/2 t2=1/√2 t3 = - 1/√2

t1=√3/2

sinx=√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=0

2x=π/2+πn, n∈Z

x=π/4+πn/2, n∈Z