If cos x

(i) $\cos x=-\frac{3}{5}$

Using the identity $\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$, we get

$\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}$

$\Rightarrow-\frac{3}{5}=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow 1-\frac{3}{5}=2 \cos ^{2} \frac{x}{2}$

$\Rightarrow \frac{2}{5}=2 \cos ^{2} \frac{x}{2}$

$\Rightarrow \frac{1}{5}=\cos ^{2} \frac{x}{2}$

$\Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{1}{5}}$

It is given that $x$ lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{5}}$

Again,

$\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}$$\Rightarrow-\frac{3}{5}=\left(-\frac{1}{\sqrt{5}}\right)^{2}-\sin ^{2} \frac{x}{2}$

$\Rightarrow-\frac{3}{5}=\frac{1}{5}-\sin ^{2} \frac{x}{2}$

$\Rightarrow-\frac{1}{5}-\frac{3}{5}=-\sin ^{2} \frac{x}{2}$

$\Rightarrow \frac{4}{5}=\sin ^{2} \frac{x}{2}$

$\Rightarrow \sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}}$

It is given that $x$ lies in the third quadrant. This means that $\frac{x}{2}$ lies in the second quadrant.

$\therefore \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$

Now,

$\sin x=\sqrt{1-\cos ^{2} x}$

$\Rightarrow \sin x=\sqrt{1-\left(-\frac{3}{5}\right)}^{2}$

$\sin x=\sqrt{1-\frac{9}{25}}=\pm \frac{4}{5}$

Since x lies in the third quadrant, sinx is negative.

$\therefore \sin x=-\frac{4}{5}$

$\Rightarrow \sin 2 x=2 \sin x \cos x$

$\Rightarrow \sin 2 x=2 \times\left(-\frac{4}{5}\right) \times\left(-\frac{3}{5}\right)$

$\Rightarrow \sin 2 x=\frac{24}{25}$

(ii) $\cos x=-\frac{3}{5}$

$\ln x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\left(\frac{-3}{5}\right)}$

$\Rightarrow \sin x=\pm \frac{4}{5}$

Here, x lies in the second quadrant.

$\therefore \sin x=\frac{4}{5}$

We know,

sin2x = 2sinx cosx

$\therefore \sin 2 x=2 \times \frac{4}{5} \times\left(-\frac{3}{5}\right)=-\frac{24}{25}$

Now,

$\cos x=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow 2 \sin ^{2} \frac{x}{2}=1-\left(-\frac{3}{5}\right)=\frac{8}{5}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{4}{5}$

$\Rightarrow \sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}}$

Since $x$ lies in the second quadrant, $\frac{x}{2}$ lies in the first quadrant.

$\therefore \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$