y = 2t^2 + t - 1 = 0. Call sin x = t. Solve the quadratic equation.
Since a - b + c = 0, then one real root is (-1) and the other is # (-c/a = 1/2).#
#Solve sin x = -1 -> x = (3pi)/2#
#Solve sin x = 1/2 -> x = pi/6#
This answer would be enough if you were to find the solutions in a given range (for example when #x in <0;2pi>#) If you are not given such range you must give all solutions remembering that trigonometric functions are periodical.
So it would be
#x = (3pi)/2 +2kpi# where #k in ZZ#
and
#x = pi/6 +2kpi# where #k in ZZ#
Check:
#x = (3pi)/2 -> y = 2 - 1 - 1 = 0# Correct
#x = pi/6 -> y = 1/2 + 1/2 - 1 = 0 # Correct.
