How do you find the x intercepts for $y = {sec}^{2} x - 1$ $y=sec^2x-1$ ?

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How do you find the x intercepts for $y = {sec}^{2} x - 1$ $y=sec^2x-1$ ?

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Answer 1 · 2016-08-01T20:17:27

$π + π n = x$ $pi+pin=x$

Explanation:

Note that ${sec}^{2} x - 1 = {tan}^{2} x$ $sec^2x-1=tan^2x$ . Hence, we can replace the original equation for a simpler one. Albeit working with $sec$ $sec$ is also easy in this case, it's a good habit to work with fewer terms. (There are occasions in which this is not true, but that will with experience.)

Also, note that the only way for ${tan}^{2} x = 0$ $tan^2x=0$ is when $tan x = 0$ $tanx=0$ .
So, when is it zero? If you remember the unit circle, you can see that $tan x = \frac{o}{a}$ $tanx=o/a$ , where $o$ $o$ is the opposite side in the unit circle, and $a$ $a$ the adjacent.

Hence, you want $o = 0$ $o=0$ . And that happens when $π + π n = x$ $pi+pin=x$ , where $π$ $pi$ is any integer.
Here's a graphical representation:
graph{tanx [-10, 10, -5, 5]}

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How do you find the x intercepts for $y = {sec}^{2} x - 1$ $y=sec^2x-1$ ?

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Explanation:

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How do you find the x intercepts for $y = {sec}^{2} x - 1$ $y=sec^2x-1$ ?