Y=2Sinx2

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)