Помогите решить cos, sin не умею их решать, заранее спасибо:)

ВАРИАНТ 1

1) 2sinx + 1 = 0, 2sinx = -1,
sinx = -1/2,
x = arcsin(-1/2) +,-2pi*n,
n = 0,1,2,3...; 2pi*n- период.

2) (V3)*tg(2x/3) = 1,
tg(2x/3) = 1/V3 = V3/3,
2x/3 = arctg(V3/3) +,-2pi*n,
n = 0,1,2,3...; 2pi*n- период.
2x = 3arctg(V3/3) +,-6pi*n,
x = (3/2)*arctg(V3/3) +,-3pi*n.

3) cos3x = - (V2/)2,
3x = arccos[-(V2)/2] +,-2pi*n,
n = 0,1,2,3...; 2pi*n- период.
x = (1/3)*arccos[-(V2)/2]+,- (2/3)*pi*n.

4) (V3)*sin[3x+ pi/6] = -V6,
sin[3x+ pi/6] = -(V6)/V3 =-[V(2*3)]/V3 = -V2,
3x+pi/6 = arcsin(-V2) +,-2pi*n,
n = 0,1,2,3...; 2pi*n- период.
3x = arcsin(-V2) +,-2pi*n, - pi/6,
x = (1/3)*arcsin(-V2) +,-(2/3)*pi*n, - pi/18 =
= (1/3)*arcsin(-V2) - pi*[1/18 -,+(2n)/3] =
= (1/3)*arcsin(-V2) - pi*[1/18 -,+(12n)/18) =
= (1/3)*arcsin(-V2) - (pi/18)*[1 -,+12n].

5) (1+cos2x)(tgx - V3) = 0,
a) 1+cos2x = 0,
b) tgx - V3 = 0.

a) cos2x = -1, 2x = arccos(-1) +,-2pi*n,
n = 0,1,2,3...; 2pi*n- период.
x = (1/2)*arccos(-1) +,-pi*n.

b) tgx = V3,
x = arctg(V3)+,- 2pi*n,
n = 0,1,2,3...; 2pi*n- период.

ВАРИАНТ 2 решается аналогично!