# Trigonometryintercepts of sine wave

#### karush

##### Well-known member
$y=\sin{2(x-\frac{\pi}{4})}$

how do you find the x intercepts of this

thot if 0=sin(2x-(pi/2)) then 0=2x-(pi/2) since sin(0)=0 but doen't look it

still don't know how to convert this to latex

K

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#### Jameson

Staff member
You're on the right track. Start with y=0 and then like you said use the fact that sin(0)=0. So solve $$\displaystyle 2x-\frac{\pi}{2}=0$$. That is only one of possibly infinite intercepts though. For what values of theta, does $$\displaystyle \sin(\theta)=0$$? Not just at 0. How can you generalize these?

• karush

#### karush

##### Well-known member
well i did this 0=2(x-(pi/4)) so from this looks like if x = (pi/4) then the intercepts this plus K(pi/4)+pi ????

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#### Ackbach

##### Indicium Physicus
Staff member
Not quite sure I would agree with the solutions so far. $\sin(z)=0$ when $z=n\pi$, for any integer $n$. So, set $2(x-\pi/4)=n\pi$ and solve for $x$. What do you get?

• Jameson

#### SuperSonic4

##### Well-known member
MHB Math Helper
I would use Ackbach's method to solve this. We know that $\sin(x) = 0$ when $x = n\pi$ where n is an integer (you can check this by graphing the result and using the periodicity of sin(x) to extend it)
Thus $2\left(x- \frac{\pi}{4}\right) = n\pi \Leftrightarrow x - \frac{\pi}{4} = \frac{n\pi}{2}$

Add $\frac{\pi}{4}$ to both sides to get the general set of solutions.

Typically these questions ask for solutions between $0$ and $2\pi$. If this is the case then $n=0,1,2$

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To render latex, use a single dollar sign at each end of your latex (or a double dollar sign to centre it, again at each end). This thread has more information including how to decode someone else's (although I go with the quote the post and look for myself method )

Code:
$2\left(x- \frac{\pi}{4}\right) = n\pi \Leftrightarrow x - \frac{\pi}{4} = \frac{n\pi}{2}$
becomes $2\left(x- \frac{\pi}{4}\right) = n\pi \Leftrightarrow x - \frac{\pi}{4} = \frac{n\pi}{2}$

• Jameson

#### Jameson

Doh! Sorry guys and thanks for catching my mistake. Clearly it's not $$\displaystyle 2 \left(x-\frac{\pi}{4} \right) + n\pi$$ but when $$\displaystyle 2 \left(x-\frac{\pi}{4} \right) = n\pi$$