$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$

It's correct. The signs are necessary because $\cos x=\frac 27$ doesn't tell us which quadrant (either 1st or 4th) the argument is in. You can also use the identity $$1+\tan^2x=\frac{1}{\cos^2 x}\Rightarrow \tan x=\pm\sqrt{\frac{1}{\cos^2 x}-1} $$ to directly compute $\tan x$ from $\cos x$.