a) sin2x = 0

x = 0+πn

x = πn\2

b) cos(x+π\4) = -1

x-π\4 = π+2πn

x = π+2πn+π\4 

x = 5π\4+2πn

g) sin(9π\2+x) = -1
9π\2+x = -π\2+2πn
x = -π\2+2πn-9π\2 
x = -5π+2πn 

v) sin(5π+x) = 1

5π+x = π\2+2πn
x = π\2+2πn-5π  
x = -9π\2+2πn   

d) 4cos(2x-π\4) = 0

cos(2x-π\4) = 0  

2x-π\4 = π\2+πn

2x = π\2+πn+π\4

2х = 3π\4+πn

х = 3π\8+πn\2

e) 1\7sin(x+3π\2) = 1

sin(x+3π\2) = 7

sin≠7 так как -1<sinA<1, ⇒ нет решений

=)..€∫∫