Question

Помогите решить уравнение!
2sinx+1=0
4cos(3x-п/4)= -√8
√3tg2/3x-1=0
ctg(x+п/6)=1/3
2sin^2x-9cosx+3=0
(1+cosx2)(tgx-√3)=0

Answer 1

1) sin x = -1/2; x = (-1)^n*(-Π/6)+Π*n
2) cos(3x-Π/4) = √2/2
3x-Π/4=Π/4+2Π*k; x1=Π/6+2Π/3*k
3x-Π/4=-Π/4+2Π*n; x2=2Π/3*n

3) tg(2x/3)=1/√3
2x/3=Π/6+Π*k; x=Π/4+Π/2*k
4) ctg(x+Π/6)=1/3; x=-Π/6+arcctg(1/3)+Π*k
5) 2sin^2 x - 9cos x + 3 = 0
2 - 2cos^2 x - 9cos x + 3 = 0
2cos^2 x + 9cos x - 5 = 0
(2cos x - 1)(cos x + 5) = 0
cos x = -5; решений нет
cos x = 1/2; x = ±Π/3 + 2Π*k

6) (1 + cos x)(tg x - √3) = 0
cos x = -1; x1 = Π + 2Π*k
tg x = √3; x = Π/3 + Π*n