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
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