<s>[INST] check my work on this problem: given tan(2x), find sin x + cos x? $\tan 2x = - 24/7$
$90^\circ < x < 180^\circ$.
Find the value of $\sin x+\cos x$.
What I have so far: 
$\tan(2x) = -\frac{24}{7} \Rightarrow \frac{2\tan(x)}{1-\tan^{2}x} = -\frac{24}{7}$.
Cross-multiplying gives us the quadratic $12\tan^{2}x-7\tan(x)-12 = 0$, factorized, it is $(3\tan x-4)(4\tan x+3) \Rightarrow \tan x = \frac{4}{3}, -\frac{3}{4}$. Since $90^\circ < x < 180^\circ$, I would have to pay attention to $\tan x = -\frac{3}{4}$. 
$\sin(\arctan(-\frac{3}{4}))+\cos(\arctan(-\frac{3}{4}))$. When calculating this, I got $\frac{1}{5}$, but that was incorrect. Where did I make my mistake?
 [/INST] You are very close - basically correct in fact. The only issue I see is that in your calculation of $\sin\arctan\left(-\frac{3}{4}\right)$ and $\cos\arctan\left(-\frac{3}{4}\right)$, you forgot the angles needed to be in the range $90^\circ$ to $180^\circ$, so you need a positive $\sin$ and negative $\cos$. If you just plug it into a calculator, you will get the wrong answer because the standard range on $\arctan$ is $-90^\circ$ to $90^\circ$.
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