How do you solve #2cos^2(theta) + sin(theta) = 1#?

#f(t) = 2cos^2 t + sin t - 1 = 0#
Replace #2cos^2 t# by #2(1 - sin^2 t)# -->
#f(t) = 2 - 2sin^2 t + sin t - 1 = 0#
#f(t) = - 2sin^2 t + sin t + 1 = 0#
Solve this quadratic equation for sin t.
Since a + b + c = 0, use shortcut. Two real roots:
sin t = 1 and #sin t = c/a = - 1/2#
Use trig table and unit circle -->
a. sin t = 1 --># t = pi/2#
b. #sin t = - 1/2# --> There are 2 solution arcs t.
#t = - (5pi)/6# or #t = (7pi)/6# --> co-terminal arcs
and #t = pi - (-(5pi)/6)= (11pi)/6.#
Answers for #(0, 2pi):#
#pi/2, (7pi)/6 and (11pi)/6#