I have a question in my further maths book to find the general solution in radians of:

$\displaystyle

\sin(2\theta + \frac{\pi}{4} ) = 1

$

The textbook says the answer is:

$\displaystyle n\pi + \frac{\pi}{8}$

Though this is what I did and got a different answer:

$\displaystyle \alpha = \arcsin(1) = \frac{\pi}{2} $

Put it into the general solution formula:

$\displaystyle 2\theta + \frac{\pi}{4} = n\pi +(-1)^n \frac{\pi}{2} $

Subtracted:

$\displaystyle 2\theta = n\pi +(-1)^n \frac{\pi}{2} - \frac{\pi}{4} $

and then divided to get:

$\displaystyle \theta = n \frac{\pi}{2} + (-1)^n \frac{\pi}{4} - \frac{\pi}{8} $

Have I gone wrong or is there a misprint?