How do you find the x intercepts of $y = sin (\frac{π x}{2}) + 1$ $y=sin((pix)/2)+1$ ?

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Answer 1 · 2017-03-17T08:53:22

$x$ $x$ -intercepts for $y = sin (\frac{π x}{2}) + 1$ $y=sin((pix)/2)+1$ are at ${.......,-5,-1,3,7,11,.......}$

To determine $x$ -intercepts of any function $y=f(x)$ , we put $y=0$

(and we put $x=0$ to find $y$ -intercepts).

Hence $x$ -intercepts are given by $f(x)=0$

and here $sin((pix)/2)+1=0$

or $sin((pix)/2)=-1=sin((3pi)/2)$

Therefore $(pix)/2=2npi+((3pi)/2)$ , where $n$ is an integer.

This happens as in the domain $0 < x < 2pi$ , only for $sin((3pi)/2)=-1$ and sine ratio has a cycle of $2pi$ .

Now as $(pix)/2=2npi+((3pi)/2)$ , and dividing by $pi$ we have

$x/2=2n+3/2$ and $x=4n+3$

Hence, we have $x$ -intercepts for $y=sin((pix)/2)+1$

at ${.......,-5,-1,3,7,11,.......}$
graph{sin((pix)/2)+1 [-10, 10, -5, 5]}

How do you find the x intercepts of y=sin((pix)/2)+1y=sin(πx2)+1y=sin((pix)/2)+1?