Thread: General Solution to sin function

1. General Solution to sin function

I have a question in my further maths book to find the general solution in radians of:
$
\sin(2\theta + \frac{\pi}{4} ) = 1
$

The textbook says the answer is:
$n\pi + \frac{\pi}{8}$

Though this is what I did and got a different answer:
$\alpha = \arcsin(1) = \frac{\pi}{2}$
Put it into the general solution formula:
$2\theta + \frac{\pi}{4} = n\pi +(-1)^n \frac{\pi}{2}$
Subtracted:
$2\theta = n\pi +(-1)^n \frac{\pi}{2} - \frac{\pi}{4}$
and then divided to get:
$\theta = n \frac{\pi}{2} + (-1)^n \frac{\pi}{4} - \frac{\pi}{8}$

Have I gone wrong or is there a misprint?

2. No, you're right.

But if $n=2p$, then $\theta=p\pi+\frac{\pi}{4}-\frac{\pi}{8}=p\pi+\frac{\pi}{8}$

and if $n=2p+1$, then $\theta=p\pi+\frac{\pi}{2}-\frac{\pi}{4}-\frac{\pi}{8}=p\pi+\frac{\pi}{8}$

So, $\theta=k\pi+\frac{\pi}{8}, \ k\in\mathbb{Z}$

3. Sorry, what does p represent?
And how does $(-1)^n$ disappear?

4. Originally Posted by alexgeek
Sorry, what does p represent?
And how does $(-1)^n$ disappear?
p is an integer.

n can be even or odd. If n is even, then n=2p and $(-1)^n=(-1)^{2p}=1$.

If n is odd then n=2p+1 and $(-1)^n=(-1)^{2p+1}=-1$

5. Oh I see you split it into two formulas to get rid of the plus/minus.
Thankyou!