The equation is given to be:
[tex]\sec\theta=-\sqrt{2}[/tex]
Recall that sec is the inverse of cos. Thus, we have:
[tex]\frac{1}{\cos\theta}=-\sqrt{2}[/tex]
Rewriting the equation, we have:
[tex]\cos\theta=-\frac{1}{\sqrt{2}}[/tex]
We can find the arccos of both sides:
[tex]\theta=\arccos(-\frac{1}{\sqrt{2}})[/tex]
Since we know that:
[tex]\cos(-x)=\cos(x)[/tex]
Then, we have:
[tex]\theta=\arccos(\frac{1}{\sqrt{2}})[/tex]
Recall the identity:
[tex]\arccos(\frac{1}{\sqrt{2}})=\frac{3\pi}{4}+2\pi n,\:θ=\frac{5\pi}{4}+2\pi n[/tex]
Therefore, the answer is the SECOND OPTION.
