## Precalculus (6th Edition) Blitzer

We have to solve the equation on the interval $[0,2\pi )$; follow the course of action given below: \begin{align} & 4{{\sec }^{2}}x-2=0 \\ & 4{{\sec }^{2}}x=0+2 \\ & {{\sec }^{2}}x=\frac{2}{4} \\ & {{\sec }^{2}}x=\frac{1}{2} \end{align} And the above expression can be further simplified as: \begin{align} & \frac{1}{{{\cos }^{2}}x}=\frac{1}{2} \\ & \text{co}{{\text{s}}^{2}}x=2 \\ & \cos x=\pm \sqrt{2} \end{align} The value of $\cos x$ cannot exceed 1. In this expression $\cos x=\pm \sqrt{2}$. Hence, it will not be a part of the solution.