4 cos2 x − 4 cos x − 1 = 0

4 cos ² x − 4 cos x − 1 = 0

$$let\;\;y=cosx\\\\
\begin{array}{rll}
4y^2-4y-1&=&0\\\\
4y^2-4y-1&=&0\\\\
\end{array}$$

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{y}} = {\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.207\: \!106\: \!781\: \!186\: \!547\: \!5}}\\
{\mathtt{y}} = {\mathtt{1.207\: \!106\: \!781\: \!186\: \!547\: \!5}}\\
\end{array} \right\}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left(-{\mathtt{0.207\: \!106\: \!781\: \!186\: \!547\: \!5}}\right)} = {\mathtt{101.952\: \!855\: \!890\: \!702^{\circ}}}$$  this is the 2nd quad solution

Other one has not solution because -1<=cosx<=1

cos x is negative so it can be in the 2nd or 3rd quadrant

1st quad equivalent = 180-101.962

$$\left({\mathtt{180}}{\mathtt{\,-\,}}{\mathtt{101.952\: \!855\: \!890\: \!702}}\right) = {\mathtt{78.047\: \!144\: \!109\: \!298}}$$   

$$\left({\mathtt{180}}{\mathtt{\,-\,}}{\mathtt{101.952\: \!855\: \!890\: \!702}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{180}} = {\mathtt{258.047\: \!144\: \!109\: \!298}}$$    this is the 3rd quad soln

$$x\approx 180(2n-1)\pm\;78.047^0 \qquad \mbox{Where }n\in\;\;\mbox{integers}$$ 

I think that looks alright.