In radians:
To get 0, only one part of the multiplication needs to be zero for the whole thing to be zero. Thus we get 2 equations:
t=0, and cos(t)=0
for the first equation, it's pretty self-explanatory and sp we get 0 as one solution.
For the second equation, cos(pi/2) and cos(3pi/2) both equal 0 so both pi/2 and 3pi/2 are answers. Since there is no bound to this problem, the actual answer is:
(2Z+1)pi/2 and 0
(2Z+1) means all odd integers
