Solve this

Given: $\cos x=\frac{-3}{5}$

To find: $\sin 2 x$

We know that,

sin2x = 2 sinx cosx …(i)

Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx

We know that

$\cos ^{2} x+\sin ^{2} x=1$

Putting the values, we get

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

$\Rightarrow \frac{9}{25}+\sin ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\frac{9}{25}$

$\Rightarrow \sin ^{2} x=\frac{25-9}{25}$

$\Rightarrow \sin ^{2} x=\frac{16}{25}$

$\Rightarrow \sin x=\sqrt{\frac{16}{25}}$

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

It is given that $\pi

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

Putting the value of sinx and cosx in eq. (i), we get

sin2x = 2sinx cosx

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

$\therefore \sin 2 \mathrm{x}=\frac{24}{25}$