Question

sec2theta-1=0

Answer 1

\[\sec^2(\theta) -1 =0\]You can treat this like a quadratic so just factor it out. It fits the difference of squares formula: \((a^2 -b^2) = (a+b)(a-b)\)

Answer 2

So you would get \[(\sec(\theta) -1)(\sec(\theta)+1) = 0\]

Answer 3

Then you want to solve each equation separately, so you'll have \[\sec(
\theta) -1=0\]\[\sec(\theta) +1=0\] and now you'll be solving for both equations

Answer 4

\[\sec(\theta) = 1 \implies \frac{1}{\cos(\theta)} = \frac{1}{1} \implies \cos(\theta) = 1\]\[\theta = \cos^{-1}(1) \implies \theta = 2\pi n\]cos(theta) = 1 every full circle around the unit circle, ending up at 2pi every time.

Answer 5

When you solve for \(\sec(\theta)=-1\) in the same manner, you'll find that \(\theta\) will also equal \(2 \pi n\)

Answer 6

I'll leave the second portion for you to figure out :)