`sinx+cosx = 0`

`sinx = -cosx`

`sinx = sin(pi/2+x)`

`x = npi+(-1)^n(pi/2+x)`

`x(1-(-1)^n) = npi+(-1)^n(pi)/2`

`x = (npi+(-1)^n(pi)/2)/(1-(-1)^n)`

*So the answer is `x = (npi+(-1)^n(pi)/2)/(1-(-1)^n).` *

`sinx+cosx = 0`

`sinx = -cosx`

`sinx = sin(pi/2+x)`

`x = npi+(-1)^n(pi/2+x)`

`x(1-(-1)^n) = npi+(-1)^n(pi)/2`

`x = (npi+(-1)^n(pi)/2)/(1-(-1)^n)`

*So the answer is `x = (npi+(-1)^n(pi)/2)/(1-(-1)^n).` *