Ive got one question in which i dont know how to solve. hopefully someone can help me! it is part b which im unsure of.
A curve is such that $\displaystyle \frac{dy}{dx} = 4sin(2X - \frac{2pi}{4}) $. THe curve passes through the point $\displaystyle (\frac{\pi}{8} , 1) $

Find the equation of the curve (ive already solved it to be Y = -2cos(2x - $\displaystyle \frac{\pi}{4} $ )+ 3

b) Given that the normal to the curve at point P is parallel to the line 2y - x = 3, find the X-coordinates of point P for X is between 0 and pi