Solve this

Given: $\sin x=\frac{\sqrt{5}}{3}$

To find: $\sin 2 x$

We know that

sin2x = 2 sinx cosx …(i)

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

We know that,

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

Putting the values, we get

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

$\Rightarrow \frac{5}{9}+\cos ^{2} x=1$

$\Rightarrow \cos ^{2} x=1-\frac{5}{9}$

$\Rightarrow \cos ^{2} x=\frac{9-5}{9}$

$\Rightarrow \cos ^{2} x=\frac{4}{9}$

$\Rightarrow \cos x=\sqrt{\frac{4}{9}}$

$\Rightarrow \cos x=\pm \frac{2}{3}$

It is given that $0

$\Rightarrow \cos \mathrm{X}=\frac{2}{3}$

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

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

$\sin 2 x=2 \times \frac{\sqrt{5}}{3} \times \frac{2}{3}$

$\therefore \sin 2 x=\frac{4 \sqrt{5}}{9}$