Помогите решить: 2tg^{2}x+3tgx-2=0 2cos^{2}x-3sinxcosx+sin^{2}x=0 9sinCosx-7cos^{2}x=2sinx cos2x=2cosx-1sinx+\sqrt{3}cosx=0

1) 2tg^2 (x)+3tg (x) -2=0 tg (x)=t 2tg^2 (t)+3t-2=0 D=b^2-4ac=25 t1,2=(-b±√D) /2a t1=-2 t2=0,5 a) tg (x)=-2 => x=arctg (-2)+pi*n б) tg (x)=0,5) => x=arctg (0,5)+pi*n 4) cos (2x)=2cos (x) -1 2cos^2 (x) -1`=2cos (x) -1 2cos^2 (x) -2cos (x)=0 2cos (x)*(cos (x) -1)=0 a) cos (x)=0 => (pi/2)+pi*n б) cos (x) -1=0 => cos (x)=1 => (pi/2)+2pi*n 6) sin (7x) -sin (x)=cos (4x) 2sin (3x/2)*cos (4x)=cos (4x) 2sin (3x/2)*cos (4x) -cos (4x)=0 cos (4x)*(2sin (3x/2) -1)=0 a) cos (4x)=0 => 4x=(pi/2)+pi*n => x=(pi/8)+pi*n/4 б) 2sin (3x/2) -1=0 => 2sin (3x/2)=1 => sin (3x/2)=1/2 => 3x/2=(pi/6)+pi*n => 3x=(pi/3)+2*pi*n => x=(pi/9)+2*pi*n/3