Explanation for (sin t -1) cos t = 0

When asked to find the general solution, I split the equation into both sin t -1
=0 and cos t=0. The most general solution I can find is pi/2 + n pi. My problem is that pi/2 + n pi isn't a solution for sin t - 1=0 alone (that solution would be pi/2 + 2n pi). Is the proper answer to the equation pi/2 + n pi?

  sin(t) – 1 = 0
    sin(t) = 1
      t = (π ⁄ 2) + (2 π • n) ... repeats every 2 π

 cos(t) = 0
   t = (π ⁄ 2) + (2 π • n)
   t = (3π ⁄ 2) + (2 π • n)

Combining all the solutions:

 t = (π ⁄ 2) + ( π • n) ... you got the right answer

You are correct that this is not a general solution for: sin(t) – 1 = 0
however the product of: (sin(t) – 1) and cos(t) is zero.

The point you are missing is that the solution is the _union_ of both solutions.
cos t = 0 OR (not AND) sin t = 1 , so indeed pi/2 + n * pi is the solution.