Probably the question should say [tex]\sin x=-0.3[/tex], as [tex]\sin x[/tex] is negative for [tex]\pi<x<\dfrac{3\pi}2[/tex].
In that case, recall that
[tex]\cos^2x+\sin^2x=1\implies\cos x=\pm\sqrt{1-\sin^2x}[/tex]
Because [tex]x[/tex] lies in the third quadrant, we expect [tex]\cos x[/tex] to also be negative, so we take the negative root above. Then
[tex]\cos x=-\sqrt{1-(-0.3)^2}\approx-0.954[/tex]
