For this problem, we are provided with the following expression:
![-\sec (-x)=5\sec (x)+1](https://img.qammunity.org/2023/formulas/mathematics/high-school/19e94o9x62lkysfdnzbl10qu2mn39a9nvh.png)
We need to solve it for x over the interval [0, 2pi).
We have:
![\sec (-x)=\sec (x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/emzlsao8hf2k2u595icvxs0rhz3642puq3.png)
Therefore, we can replace the left side of the equation as shown:
![\begin{gathered} -\sec (x)=5\sec (x)+1 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/m0cfw7zupr5emqizhkd3c4q4v5rgnuizou.png)
Now we need to isolate the sec(x) on the left side.
![\begin{gathered} -5\sec (x)-\sec (x)=1 \\ -6\sec (x)=1 \\ \sec (x)=-(1)/(6) \\ (1)/(\cos (x))=-(1)/(6) \\ \cos (x)=-6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wg1hat2ouc7dowv91izdhg852s5di2rj1g.png)
Now we can apply the arc cosine to determine the value of x.
![\begin{gathered} \arccos (\cos (x))=\arccos (-6) \\ x=\arccos (-6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/visxo06h1he9wx01s5vt7ll8sm54mnej8j.png)
There are no real values for x that have a cosine equal to -6. Therefore, this problem has no real solution.