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In the interval $( - \pi ,\ \pi ]$.There are 2 roots exist mentioned in the book. Could anyone please explain how?

Exact question from book : Let $Y=\sec X$ .Compute $f_Y(y)$ in terms of $f_X(x)$ .What is $f_Y(y)$ when $f_X(x)$ is uniform in $[ −π,π )$ ?

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    $\begingroup$ Note that $\displaystyle y = \sec(x) ={1\over\cos(x)}$. The only way a fraction equals $0$ is if the numerator equals zero. This is impossible since $1\ne0$, which implies there are no real roots at all. $\endgroup$
    – Decaf-Math
    Commented Oct 16, 2018 at 6:47
  • $\begingroup$ For y >1 ,there are 2 solution exists mentioned in the book. Actually the question belongs to Random variable from Q.# 3.12 from Probability, Statistics, and Random Processes for Engineers, 4th Edition-Henry Stark and John Woods $\endgroup$
    – Pramod
    Commented Oct 16, 2018 at 6:52
  • $\begingroup$ The likelihood someone here has the textbook is low. If you want further insight, are you able to provide the context of the problem listed? Edit your post with the exact question, word-for-word. $\endgroup$
    – Decaf-Math
    Commented Oct 16, 2018 at 6:57
  • $\begingroup$ Okay, Q .Let Y=secX .Compute $ f_{Y}(y) $ in terms of $ f_{X}(x) $ .What is $ f_{Y}(y) $ when $ f_{X}(x) $ is uniform in [ $ - \pi, \pi $ ) $\endgroup$
    – Pramod
    Commented Oct 16, 2018 at 7:27

2 Answers 2

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$$\sec x = \frac{1}{\cos x}$$

Obviously it is impossible for $\sec x$ to become $0$ because $\frac{1}{\cos x}$ can never reach $0$. In fact, since the range of $\cos x$ is $y \in [-1, 1]$, the range of $\sec x$ becomes $y \in (-\infty, -1] \cup [1, +\infty)$. Hence, there are no real roots.

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Hint #1: Draw the graph (or use a graphing program) of $\sec (x) \ \text {(or} \ \dfrac {1}{\cos (x)})$.

Hint #2: Viewing the graph, where is $\sec (x)$ undefined?

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  • $\begingroup$ Are you meant solutions are $x1=cos^{-1} \frac{1} {y} $ and $ x2= -cos^{-1} \frac{1} {y} $ $\endgroup$
    – Pramod
    Commented Oct 16, 2018 at 13:28
  • $\begingroup$ I've re-written the hints above. You may also want to see the solution offered by KM101 below. $\endgroup$
    – bjcolby15
    Commented Oct 16, 2018 at 14:00
  • $\begingroup$ My book solution is something i mentioned above . $\endgroup$
    – Pramod
    Commented Oct 16, 2018 at 14:20

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