Answer
The solution is $\underline{x=1.3770,4.9062}.$
Work Step by Step
We know that for a standard quadratic equation of the form $a{{x}^{2}}+bx+c=0$, the solution is:
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
And replace $x\text{ by }\cos x\text{ }$ in the standard quadratic equation to obtain the given equation in $\cos x$. Then the solution of the given equation is:
$\begin{align}
& \cos x=\frac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-\left( 4\times 1\times -1 \right)}}{2\times 1} \\
& \cos x=\frac{-5\pm \sqrt{25+4}}{2} \\
\end{align}$
After that, simplify the equation:
$\begin{align}
& \cos x=\frac{-5+\sqrt{29}}{2} \\
& \cos x\approx 0.1926 \\
\end{align}$
And the solution for addition will be:
$\begin{align}
& \cos x=\frac{-5-\sqrt{29}}{2} \\
& \cos x\approx -5.1926 \\
\end{align}$
We know that $\operatorname{cosine}$ of any number lies only in the interval $\left[ -1,1 \right]$. So $\cos x\ne -5.1926$ and it is not the solution. The only solution is:
$\begin{align}
& \cos x\approx 0.1926 \\
& x\approx {{\cos }^{-1}}\left( 0.1926 \right)
\end{align}$
We have to calculate the equation using a calculator; it is required to keep the calculator in radian mode and find the value of ${{\cos }^{-1}}\left( 0.1926 \right)$
$\begin{align}
& x\approx {{\cos }^{-1}}\left( 0.1926 \right) \\
& \approx 1.3770
\end{align}$
And the cosine value is positive in I and IV quadrants. So
$x\approx 1.3770$
Thus, the other solution will be:
$\begin{align}
& x\approx 2\pi -1.3770 \\
& \approx 2\times 3.14159-1.3770 \\
& \approx 4.9062
\end{align}$
