0= 3cos(π/2 x +2π)

0= cos(π/2 x +2π)

cos(x) is 0 when x= π/2 and 3π/2

so lets say that (π/2 x +2π) has to equal π/2, 3π/2, -π/2,or -3π/2

i'll show you π/2, 3π/2 you can do -π/2,or -3π/2

π/2 x +2π = π/2 ............ and......... π/2 x +2π = 3π/2

......π/2 x = - 3π/2 ............................... π/2 x = -π/2

............x = -3 .......................................... x = -1

if you worked out the other two problems you would get that the particle is at rest when x (or t) is equal to -1 and -3... and -5 and 1

you can also find the answer by graphing it on graphing calculator and calculating the "zeros"