Помогите решить:

2tg^{2}x+3tgx-2=0

2cos^{2}x-3sinxcosx+sin^{2}x=0

9sinCosx-7cos^{2}x=2sinx

cos2x=2cosx-1

sinx + / sqrt{3}cosx=0

sin7x-sinx=cos4x

это задание уже было сегодня

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