If [tex]\pi<\theta<\dfrac{3\pi}2[/tex], then [tex]\cos\theta<0[/tex]. By the Pythagorean theorem, we then have
[tex]\sin^2\theta+\cos^2\theta=1[/tex]
[tex]\implies\cos\theta=-\sqrt{1-\sin^2\theta}[/tex]
[tex]\implies\cos\theta=-\dfrac{2\sqrt2}3[/tex]
