how would you go about solving this:
if sin x = -4/5, and 270 degrees < x < 360 degrees, then find tan2x
$\displaystyle \displaystyle\tan 2x=\frac{2\tan x}{1-\tan^2x}$ (1)
If $\displaystyle x\in (270^{\circ},360^{\circ})\Rightarrow \cos x>0$.
Then, $\displaystyle \cos x=\sqrt{1-\sin^2x}=\frac{3}{5}\Rightarrow\tan x=\frac{\sin x}{\cos x}=-\frac{4}{3}$.
Now, substitute $\displaystyle \tan x$ in (1) and find $\displaystyle \tan 2x$.