x-intercept is where y=0. Therefore we have the following equation:

`sin(pi x)+cos(pi x)=0`

Now we square the whole equation.

N.B. Sometimes when we square the whole equation we get some solutions that are not solutions to the original equation so we need to check them first. Fortunately this is not the case here.

`sin^2(pi x)+2sin(pi x)cos(pi x)+cos^2(pi x)=0`

Now we use Pythagorean trigonometric identity `sin^2 x+cos^2 x=1.`

`2sin(pi x)cos(pi x)+1=0`

Now we use formula for sine of double angle `sin(2x)=2sin x cos x`

`sin(2pi x)=-1`

Sine is equal to -1 for `(3pi)/2+2k pi,\ k in ZZ.`

`2pi x=(3pi)/2+2k pi`

`x=3/4+k,\ k in ZZ`

Now we check the solution for `k=0` and `k=1,` because the functions basic period is 2 (not 1).

`sin((3pi)/4)+cos((3pi)/4)=sqrt2/2-sqrt2/2=0` First solution is OK.

`sin((7pi)/4)+cos((7pi)/4)=-sqrt2/2+sqrt2/2=0` Second solution is OK.

**x-intercepts of the graph are** `x=3/4+k,\ k in ZZ.`

You can also solve this by using tangent half-angle substitution.