Answer
The solutions to the given equation are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\ \text{ and }\ \frac{7\pi }{4}$.
Work Step by Step
We have to solve the equation on the interval $[0,2\pi )$:
$\begin{align}
& 5\text{se}{{\text{c}}^{2}}x-10=0 \\
& \text{se}{{\text{c}}^{2}}x=2
\end{align}$
By using the reciprocal identity of trigonometry ${{\sec }^{2}}x=\frac{1}{{{\cos }^{2}}x}$.
$\begin{align}
& \frac{1}{{{\cos }^{2}}x}=2 \\
& \text{co}{{\text{s}}^{2}}x=\frac{1}{2} \\
& \text{cos}\ x=\sqrt{\frac{1}{2}}
\end{align}$
Multiply 2 with the numerator and the denominator of the given expression:
$\begin{align}
& \text{cos}\ x=\sqrt{\frac{1}{2}\times \frac{2}{2}} \\
& =\sqrt{\frac{2}{4}} \\
& =\frac{\sqrt{2}}{2}
\end{align}$
After that, simplify it by solving the equation of $\frac{\sqrt{2}}{2}$.
And take the expression both as a negative and positive value.
$\text{cos}\ x=\pm \frac{\sqrt{2}}{2}$
Each factor needs to be calculated as:
$\text{cos}\ x=\frac{\sqrt{2}}{2}$
Or
$\text{cos}\ x=-\frac{\sqrt{2}}{2}$
So, in the quadrant graph, the value of cosine is $\frac{\sqrt{2}}{2}$ at $\frac{\pi }{4}$ , and $\frac{7\pi }{4}$. This implies,
$\begin{align}
& \text{cos}\ x=\cos \frac{\pi }{4} \\
& x=\frac{\pi }{4}
\end{align}$
Then, take the value of $x$ as $\frac{7\pi }{4}$:
$\begin{align}
& \text{cos}\ x=\cos \frac{7\pi }{4} \\
& x=\frac{7\pi }{4}
\end{align}$
And,
So, in the quadrant graph, the value of cosine is $-\frac{\sqrt{2}}{2}$ at $\frac{3\pi }{4}$ , and $\frac{5\pi }{4}$. This implies,
$\begin{align}
& \text{cos}\ x=\cos \frac{3\pi }{4} \\
& x=\frac{3\pi }{4}
\end{align}$
Now, take the value of $x$ as $\frac{5\pi }{4}$:
$\begin{align}
& \text{cos}\ x=\cos \frac{5\pi }{4} \\
& x=\frac{5\pi }{4}
\end{align}$
These are the proposed solutions of the tangent functions.
Thus, the actual solution in the interval $[0,2\pi )$ will be $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\ \text{ and }\ \frac{7\pi }{4}$.
