<s>[INST] $50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$ 
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$

I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\dfrac{\sqrt{55}}{3}$
Are these $\pm$ signs correct? My reasoning came for a CAST diagram, can you confirm if this is correct and if not, why?
 [/INST] 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 $\cos x$.
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