Question

2⋅ (cos(x))²= 3⋅ sin(x)
x ∈ [ 0; 2π [

Answer 1

$2cos^2x=3sinx \ 2-2sin^2x-3sinx=0 \ -2sin^2x-3sinx+2=0 \ sinx=t, t=[-1,1] \ -2t^2-3t+2=0 \ D=9+16 \ sqrt{D} =5 \ t_{1} = frac{3+5}{-4}=-2 neq root neq [-1;1]; \ t_{2} =1/2 \ sinx=1/2 \ x=(-1)^n frac{pi }{6} + pi n, k=Z$
Отбор корней: $0 leq x=(-1)^nfrac{pi }{6} + pi nleq 2 pi \ n=0, x= frac{ pi }{6} =root \ n=1, x= frac{-7 pi }{6} neq root \ n=-1, x= frac{5 pi }{6}=root \ n=2, x= frac{13 pi }{6} neq root, \ n=-2, x neq root \ End.$
Ответ: $frac{ pi }{6} , frac{5 pi }{6}$