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}#
