[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}<0[/tex]
[tex]\sin\theta=\dfrac23>0[/tex], which tells us that [tex]\cos\theta[/tex] must be negative.
Recall that
[tex]\sin^2\theta+\cos^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}[/tex]
where we take the negative square root because we know to expect [tex]\cos\theta<0[/tex]. We find that
[tex]\cos\theta=-\sqrt{1-\dfrac49}=-\dfrac{\sqrt5}3[/tex]
