Question:
Find the solution of the equation if it is in the first quadrant
$\displaystyle sec^2x-1=0 $
Attempt:
$\displaystyle secx=1 ; x= \frac {1}{arccos 1} $
I keep getting infiniate as my answer
Thank you
$\displaystyle \sec^2{x} = 1$
$\displaystyle \frac{1}{\cos^2{x}} = 1$
$\displaystyle 1 = \cos^2{x}$
$\displaystyle \pm 1 = \cos{x}$
If $\displaystyle \cos{x} = -1$ then $\displaystyle x = \{\dots -3\pi, -\pi, \pi, 3\pi, \dots\}$ and if $\displaystyle \cos{x} = 1$ then $\displaystyle x = \{ \dots, -2\pi, 0, 2\pi, \dots \}$.
Therefore $\displaystyle x = n\pi$ where $\displaystyle n$ is an integer.