Find value of $\sin x-\frac{1}{\cot x}$ If $\sin x+\frac{1}{\cot x}=3$, calculate the value of $\sin x-\frac{1}{\cot x}$

Please kindly help me

Let $\sin x -\frac{1}{\cot x}=t$
Then, $$\sin x= \frac{3+t}{2}, \cot x= \frac{2}{3-t}$$
By using $$1+\cot ^2x= \frac{1}{\sin^2 x}$$
Then, the equation $$t^4-18t^2+48t+81=0$$

Hint:
$\sin x+\dfrac1{\cot x}=3$
Let $\sin x-\dfrac1{\cot x}=y$
Solve for $\sin x, \dfrac1{\cot x}$
Now use $$\dfrac1{\sin^2x}-\cot^2x=1$$
