[sinx+cos(π+x)]/[sinx+sin(π/2-x)]
=(sinx-cosx)/(sinx+cosx)
=(tanx-1)/(tanx+1)
=1-[2/(tanx+1)]
sinx=√3/3
∴cosx=±√6/3
∴tanx=±√2/2
∴原式=1-{2/[(√2/2)+1]}=2√2-3
或原式=1-{2/[(-√2/2)+1]}=-2√2-3
即原式=±2√2-3