Question: Solving trigonometric equations for a particular set of solutions rather than one or AllSolutions

0 = 0, -(1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4)*(-Pi*x3+4*cos(Pi*t)) = 0, -(1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4)*(-Pi*x2+4*sin(Pi*t)) = 0, (1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4) = 0, 0 = 0, -(1/8)*cos(l1)*cos(l3)*cos(l4)*(-Pi*x3+4*cos(Pi*t)) = 0, -(1/8)*cos(l1)*cos(l2)*cos(l4)*(-Pi*x2+4*sin(Pi*t)) = 0, (1/8)*cos(l1)*cos(l2)*cos(l3) = 0

After I ran solve({eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8}, {l1, l2, l3, l4, t, x1, x2, x3}, AllSolutions); I got the general solutions but I need only one particular subset of all solutions;

A much smaller version of my question is for

solve(cos(x)=0,AllSolutions) gives (1/2)*Pi+Pi*_Z297

I gave the 'AllSolutions' option to get the value of -pi/2 along with pi/2(which is the only solution given if we do not give AllSolutions option). Simple substitution for _Z297 = -1 would give -pi/2 but this constant is arbitary and also when equations count is more than one, it is becoming very difficult to substitute and do the work. 

Can anyone suggest a work around or procedure to get both pi/2 and -pi/2 from the general set of solutions.

Thank you,

Pradeep.