[latex]3ctgx+2sinx=0[/latex] Помогите решить уравнение

3ctgx + 2sinx = (3cosx + 2 sin^2x)/sinx = 0sinx <> 03cosx + 2(1 - cos^2x) = 0замена y = cosx2y^2 - 3y - 2 = 0D = 9 + 16 = 25y1 = (3 - 5)/4 = - 0,5y2 = (3 + 5)/4 = 2   - не подходит,  > 1cosx = -0,5x = - pi/3 + 2pin ,  n ∈ Z

[latex]3ctgx+2sinx=0\ frac{3cosx}{sinx} +2sinx=0\ \ frac{3cosx+2sin ^{2} x}{sinx} =0 \ sinx eq 0 \ x eq pi n \ \ 3cosx+2sin ^{2} x=0\2-2cos ^{2} x+3cosx=0 \ -2cos ^{2} x+3cosx+2=0 \ cosx=t\-1 leq t leq 1 \ -2t ^{2} +3t+2=0\ [/latex][latex]2t ^{2} -3t-2=0\D= 9+16=25 \ sqrt{D} =5 \ t _{1} = frac{3+5}{4} = frac{8}{4} =2 \ \ t_{2} = frac{3-5}{4} =- frac{1}{3} [/latex][latex]cosx=- frac{1}{2} \ \ x= frac{2 pi }{3} + pi k[/latex]k,n ∈ Z