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.