Precalculus (6th Edition) Blitzer

The solution on the interval $[0,2\pi )$, correct to four decimal places is $2.2370$ or $4.0462$.
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 $4.0462$.