Change to Sin(x) = Cos(x). Since Sine is y/r and Cos is x/r you know that you're looking on the unit circle where x=r, which occurs at pi/4 and 5pi/4.
Alternatively, square both sides
(sinx)^2 = (cosx)^2
change (cosx)^2 = 1 - (sinx)^2
to get:
(sinx)^2 = 1 - (sinx)^2
rearrange
(sinx)^2 = 1/2
sinx = +- root(2)/2
This occurs at pi/4, 3pi/4, 5pi/4, 7pi/4, however we have too many solutions, because both sin and cos must have same sign( extrenious solutions created by squaring and square rooting), so only at pi/4 and 5pi/4
Alternatively, multiply both sides by the conjugate (sinx + cosx).
(sinx)^2 - (cosx)^2 = 0
Use identity (sinx)^2 - (cosx)^2 = cos2x
cos2x = 0
2x = pi/2 and 3pi/2
divide all by 2
x = pi/4 and 3pi/4