[tex]\sin\theta=\dfrac13>0[/tex], so
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}<0\implies\cos\theta<0[/tex]
Recall that
[tex]\cos^2\theta+\sin^2\theta=1[/tex]
for all [tex]\theta[/tex], and knowing that [tex]\cos\theta<0[/tex] we have
[tex]\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac{2\sqrt2}3[/tex]
