Answer
The solution on the interval $[0,2\pi )$, correct to four decimal places is $2.2370$ or $4.0462$.
Work Step by Step
For a standard quadratic equation of the form $a{{x}^{2}}+bx+c=0$, the solution is:
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Replace $x\text{ by }\cos x\text{ }$ in the above formula to obtain the given equation in $\cos x$.
Then, the solution of the equation is:
$\begin{align}
& \cos x=\frac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times -1}}{2\times 1} \\
& =\frac{1\pm \sqrt{1+4}}{2} \\
& =\frac{1\pm \sqrt{5}}{2}
\end{align}$
This implies that,
$\begin{align}
& \cos x=\frac{1-\sqrt{5}}{2} \\
& =\frac{1-2.23607}{2} \\
& \approx -0.6180
\end{align}$
Or,
$\begin{align}
& \cos x=\frac{1+\sqrt{5}}{2} \\
& =\frac{1+2.23607}{2} \\
& \approx 1.6180
\end{align}$
We know that $\operatorname{cosine}$ of any number lies only in the interval $\left[ -1,1 \right]$.
Thus, $\cos x\ne 1.6180$ and it is not the solution.
The only solution is:
$\begin{align}
& \cos x\approx -0.6180 \\
& x\approx {{\cos }^{-1}}\left( -0.6180 \right)
\end{align}$
Now, to calculate the equation with the help of a calculator, we use radian mode:
$\theta ={{\cos }^{-1}}\left( 0.6180 \right)\approx 0.9046$
Since, the cosine value is negative in II and III quadrants:
$x\approx \pi -0.9046$
It is known that the value of $\pi $ is $3.14159$.
Thus, the value of $x$ is:
$\begin{align}
& x=\pi -0.9046 \\
& =3.14159-0.9046 \\
& \approx 2.2370
\end{align}$
Or,
$x\approx \pi +0.9046$
The value of $\pi $ is $3.14159$.
Thus, the value of $x$ is:
$\begin{align}
& x=\pi +0.9046 \\
& =3.14159+0.9046 \\
& \approx 4.0462
\end{align}$
Therefore, the solution of the equation ${{\cos }^{2}}x-\cos x-1=0$ on the interval $[0,2\pi )$, correct to four decimal places is $2.2370$ or $4.0462$.