<s>[INST] Solving for Cos Exactly How to solve $\cos(\dfrac{5\pi}{6})$ and  $\cos(\dfrac{7\pi}{6})$ exactly?
I couldn't use special triangles to solve this either.
 [/INST] Remember these 2 identities: $$\begin{align} &\cos \left( \pi - x \right) = - \cos(x) \\ & \cos(\pi + x ) = -\cos (x) \end{align}$$
Or you can simply write: $$\cos (\pi - x) = \cos (\pi + x) =-\cos (x) $$
You need to find $\cos \left(\cfrac{5\pi}{6}\right)$. Try to break this down in order to use the above identities. 
You can write $\cos \left(\cfrac{5\pi}{6}\right)$ as $\cos \left( \cfrac{6\pi - \pi}{6}\right)$ 
Or: $$\cos \left(\cfrac{6\pi}{6} - \cfrac{\pi}{6} \right) = \cos \left(\pi - \cfrac{\pi}{6} \right) $$
Now, you may use $\cos \left( \pi - x \right) = -\cos x$ identity here.
For the second part: $\cos \left(\cfrac{7\pi}{6}\right) = ? $ 
Break $\cfrac{7\pi}{6}$ into: $\color{blue}{\cfrac{6\pi + \pi}{6}}$ 
So, you get: 
$$\cos \left( \pi + \cfrac{\pi}{6} \right)$$ 
Thus, you can use the identity of $\cos (\pi + x) = -\cos x$ 
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