How do you find the x intercepts of #y=sinpix+cospix#?

1 Answer
Feb 20, 2018

#x = k - 1/4" "# for any integer #k#

Explanation:

Note that:

#y = sin pi x + cos pi x#

#color(white)(y) = sqrt(2)(sqrt(2)/2 sin pi x + sqrt(2)/2 cos pi x)#

#color(white)(y) = sqrt(2)(sin (pi/4) sin pi x + cos (pi/4) cos pi x)#

#color(white)(y) = sqrt(2) sin (pi/4 + pi x)#

Also note that:

#sin theta = 0" "# if and only if #theta = k pi# for some integer #k#

So we require:

#pi/4 + pi x = k pi#

Dividing both sides by #pi#, this becomes:

#1/4 + x = k#

Then subtracting #1/4# from both sides we find:

#x = k - 1/4" "# for any integer #k#