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).(−ca=12).
Solve sin x = -1 -> x = (3pi)/2Solvesinx=−1→x=3π2
Solve sin x = 1/2 -> x = pi/6Solvesinx=12→x=π6
This answer would be enough if you were to find the solutions in a given range (for example when x in <0;2pi>x∈<0;2π>) 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 +2kpix=3π2+2kπ 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.
