1)2cos в квадрате x + 5cos x=3;
2)tg x - 3ctg x=0;
3)sin 3x-sin x=0;
4)2 sin x + sin 2 x=0
Если не трудно,буду благодарен:)

1) заменим cosx = a

2a² + 5a - 3 =0

-5±√25+24 /4 = a1 = -3, а2 = 1/2

а1 не подходит.

cosx = 1/2

x = ±π/3 + 2πn

2) tgx - 3ctgx = 0, tgx - 3/tgx = 0, tg²x-3 = 0, 1-cos2x/1+cos2x = 3

4cos2x = -2, сos2x = -1/2

2x =  ±π-arccos1/2 +2πn

2x = ± π-π/3 + 2πт = 2π/3 + 2πт

x = ±π/3 + πn

3) Sin3x - Sinx = 0

2cos(3x+x/2)*sin(3x-x/2)

2cos2xsinx = 0

cos2xsinx = 0

1) cos2x = 0           2) sinx = 0

2x = π/2+πn               x = πn

x = π4+πn/2    

4) 2sinx +sin2x = 0, 2sinx +2sinxcosx = 0 , 2sinx(1+cosx) = 0 

     1) 2sinx = 0                   2) 1+cosx = 0

         sinx = 0                         cosx = -1

         x = πn                           x = π + 2πn