First of all, factorize:
color(white)(xxx) 2 sin t cos t - cos t = 0
<=>color(white)(x) cost (2 sin t - 1 ) = 0
A product can only be 0 if one of the factors (or both) is 0 which means:
<=>color(white)(x) cos t = 0 color(white)(xx) "or" color(white)(xx) 2 sin t - 1 = 0
<=>color(white)(x) cos t = 0 color(white)(xx) "or" color(white)(xx) sin t = 1/2
We know that the cosinus function intercepts with the x axis at pi/2 and 3/2 pi in the period [0, 2pi].
And we know that sin(t) = 1/2 is true for t = pi / 6 and t = (5 pi) / 6 in the period [0, 2pi].
So, in the period [0, 2pi] we have four solutions:
t = pi/6 " or " t = pi/2 " or " t = (5 pi)/6 " or " t = (3 pi)/2.
To obtain a solution space for t in RR, we need to consider all possible periods. Then, the solution is
t = pi/2 + k " or " t = pi/6 + 2k " or " t = (5pi)/6 + 2k, for k in ZZ
The solution space can be also formulated as
{ pi/2 + k; color(white)(x) pi/6 + 2k; color(white)(x)(5pi)/6 + 2k color(white)(x) | color(white)(x) k in ZZ}
