Answer
The solution is $\underline{x=0.7494,5.5338}.$
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 provide equation is:
$\begin{align}
& \cos x=\frac{-\left( 2 \right)\pm \sqrt{{{\left( 2 \right)}^{2}}-\left( 4\times 1\times -2 \right)}}{2\times 1} \\
& =\frac{-2\pm \sqrt{4+8}}{2} \\
& =\frac{-2\pm \sqrt{12}}{2}
\end{align}$
That is,
$\begin{align}
& \cos x=\frac{-2+\sqrt{12}}{2} \\
& =\frac{-2+3.4641}{2} \\
& \cos x\approx 0.7321
\end{align}$
Or,
$\begin{align}
& \cos x=\frac{-2-\sqrt{12}}{2} \\
& \cos x\approx -2.7321 \\
\end{align}$
But $\operatorname{cosine}$ of any number lies only in the interval $\left[ -1,1 \right]$. So $\cos x\ne -2.7321$ and it is not the solution. The only solution is:
$\begin{align}
& \cos x\approx 0.7321 \\
& x\approx {{\cos }^{-1}}\left( 0.7321 \right)
\end{align}$
We have to calculate the equation using a calculator; it is required to keep the calculator in radian mode and find the value of ${{\cos }^{-1}}\left( 0.7321 \right)$.
$\begin{align}
& x\approx {{\cos }^{-1}}\left( 0.7321 \right) \\
& \approx 0.7494
\end{align}$
And the cosine value is positive in I and IV quadrants. So,
$x\approx 0.7494$
And the other value will be:
$\begin{align}
& x\approx 2\pi -0.7494 \\
& \approx 2\times 3.14159-0.7494 \\
& \approx 5.5338
\end{align}$
Where, $\pi =3.14159$.
