# Thread: Need help understanding this problem..

1. ## Need help understanding this problem..

Okay, I have to find all solutions to the equation and I'm stuck.

$sin(x + pi/4) = 1/2$

I unno what to do. Should I move that pi/4 to the right?

2. Originally Posted by Megaman
Okay, I have to find all solutions to the equation and I'm stuck.

$sin(x + pi/4) = 1/2$

I unno what to do. Should I move that pi/4 to the right?
couple of hints ...

$\sin\left(\frac{\pi}{6} + 2k\pi\right) = \frac{1}{2}$

$\sin\left(\frac{5\pi}{6} + 2k\pi \right) = \frac{1}{2}$

$k$ is an integer

3. Okay, I know that there are two possible solutions to this that being 30 degrees and 330 degrees. I don't understand where that 2kpi is coming from though. Is that like the same thing as 2pi being one circle while k is how many times it should loop? Anywho, I'm still confused over this. I got the answer by subtracting pi/4 from pi/6 and 5pi/6, which would make it -pi/12 and 7pi/12.

4. Originally Posted by Megaman
Okay, I know that there are two possible solutions to this that being 30 degrees and 330 degrees. I don't understand where that 2kpi is coming from though. Is that like the same thing as 2pi being one circle while k is how many times it should loop? Anywho, I'm still confused over this. I got the answer by subtracting pi/4 from pi/6 and 5pi/6, which would make it -pi/12 and 7pi/12.
the original problem says to find all solutions for x ...

$x + \frac{\pi}{4} = \frac{\pi}{6} + 2k\pi$

$x = 2k\pi - \frac{\pi}{12}
$

$x + \frac{\pi}{4} = \frac{5\pi}{6} + 2k\pi
$

$x = \frac{7\pi}{12} + 2k\pi$

5. Originally Posted by skeeter
the original problem says to find all solutions for x ...

$x + \frac{\pi}{4} = \frac{\pi}{6} + 2k\pi$

$x = 2k\pi - \frac{\pi}{12}
$

$x + \frac{\pi}{4} = \frac{5\pi}{6} + 2k\pi
$

$x = \frac{7\pi}{12} + 2k\pi$

$\sin{\left(x + \frac{\pi}{4}\right)} = \frac{1}{2}$

$x + \frac{\pi}{4} = \arcsin{\left(\frac{1}{2}\right)}$

$x + \frac{\pi}{4} = \left\{\frac{\pi}{6},\pi - \frac{\pi}{6}\right\} + 2\pi n$ where n is an integer, since sine is positive in the first and second quadrants.

$x + \frac{\pi}{4} = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n$

$x = \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} + 2\pi n - \frac{\pi}{4}$.

Can you go from here?