[tex]\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\pm\sqrt{1-\sin^2\theta}[/tex]
Because [tex]\theta[/tex] lies in quadrant II, we can expect [tex]\cos\theta[/tex] to be negative, so we take the negative root.
This leaves us with
[tex]\cos\theta=-\sqrt{1-\left(\dfrac35\right)^2}=-\dfrac45[/tex]
