1. ## Find x-intercepts

Hi
I am having trouble finding the x-intercepts from $-2\pi to 2\pi$
from the equation: $y=2sin(x-\frac{\pi}{4})+1$

This is what i have done but it doesn't match the book's answers:

$0=2sin(x-\frac{\pi}{4})+1$
$\frac{-1}{2}=sin(x-\frac{\pi}{4})$
$x-\frac{\pi}{4}=\pi+\frac{\pi}{6},2\pi-\frac{\pi}{6},-\pi+\frac{\pi}{6},-2\pi-\frac{\pi}{6}$
$x=\frac{7\pi}{6}+\frac{\pi}{4},\frac{11\pi}{6}+\fr ac{\pi}{4},\frac{-5\pi}{6}+\frac{\pi}{4},\frac{-13\pi}{6}+\frac{\pi}{4}$
$\frac{17\pi}{12},\frac{25\pi}{12},\frac{-7\pi}{12},\frac{17\pi}{8}$

P.S

2. Whilst your answers are technically correct, 2 of them do not satisfy the limits. This is because your limits were wrong when solving $\sin(x - \frac{\pi}{4}) = \frac{-1}{2}$

Be careful when you are solving questions like this. If the angle is $\theta + k$, the limits you need to apply when solving for $\theta + k$ changes. If you had limits $-2\pi \le \theta \le 2\pi$, the limits for $\theta + k$ should be $-2\pi + k \le \theta + k \le 2\pi + k$

3. So it should be like this:
$\pi+\frac{\pi}{6},-\pi+\frac{\pi}{6},2\pi+\frac{\pi}{6},-2\pi+\frac{\pi}{6}$

However for the last two i get the incorrect answers.

4. How did you figure the last two solutions?

General solution for $x - \frac{\pi}{4}$ is:

$x - \frac{\pi}{4} = n\pi + (-1)^n \left(\frac{-\pi}{6}\right)$

Limits: $-2\pi - \frac{\pi}{4} \le x - \frac{\pi}{4} \le 2\pi - \frac{\pi}{4}$

For $n = 0 \Rightarrow \left(\frac{-\pi}{6}\right)$
For $n = 1 \Rightarrow \pi - \left(\frac{-\pi}{6}\right) = \pi + \left(\frac{\pi}{6}\right)$
For $n = -1 \Rightarrow -\pi - \left(\frac{-\pi}{6}\right) = -\pi + \left(\frac{\pi}{6}\right)$
For $n = 2 \Rightarrow 2\pi + \left(\frac{-\pi}{6}\right) = 2\pi - \left(\frac{\pi}{6}\right)$ but this exceeds the limit, so it's discounted.
For $n = -2 \Rightarrow -2\pi + \left(\frac{-\pi}{6}\right) = -2\pi -\left(\frac{\pi}{6}\right)$

Hence $x = \frac{\pi}{12} ; \frac{17\pi}{12} ; \frac{-7\pi}{12} ; \frac{-23\pi}{12}$

5. Originally Posted by Gusbob
How did you figure the last two solutions?
yeh, i made the denominator 24 and not 12.
I tried it again with 12 and it was correct.