y = 2sin(x²) and y = 2sin(2x - π/2)
---> 2sin(x²) = 2sin(2x - π/2) ---> sin(x²) = sin(2x - π/2) ---> x² = 2x - π/2
---> x² - 2x + π/2 = 0
---> (using the quadratic formula with a = 1, b = -2, and c = π/2) ---> x = [2 + √(4 - 4(1)(π/2)]/2
---> x = [2 + √(4 - 2π)]/2 ---> non-real answer
However, since sin(x) = sin(π-x), let's change x² = 2x - π/2 to x² = π - [2x - π/2]
---> x² = -2x + 3π/2
---> x² - 2x - 3π/2 = 0
Again, using the quadratic formula: x = [2 + √(4 - 4(1)(-3π/2)]/2
---> x = [2 + √(4 +6π)]/2 ---> x = 1.39 (approx) (by symmetry, also x = -1.39)
Using a period of 2π: x² = 3π - [2x - π/2] (replace π wirh 3π)
---> x² = -2x + 7π/2
---> x² - 2x - 7π/2 = 0
Again, using the quadratic formula: x = [2 + √(4 - 4(1)(-7π/2)]/2
---> x = [2 + √(4 +14π)]/2 ---> x = 2.46 (approx) (by symmetry, also x = -2.46)