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

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Answer 1 · 2017-03-17T08:53:22

#x#-intercepts for #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]}