Question 597341
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -2\csc\left(x\ -\ \frac{\pi}{2}\right)\ +\ 4\ \ ] in *[tex \LARGE \ \ \left[0,\pi\right]]


I presume you mean the zeros of the function since "roots" is really a term that is used with polynomial functions.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\csc\left(x\ -\ \frac{\pi}{2}\right)\ +\ 4\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\csc\left(x\ -\ \frac{\pi}{2}\right)\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\left(x\ -\ \frac{\pi}{2}\right)\ =\ 2]


Cosecant is the reciprocal of sine


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\sin\left(x\ -\ \frac{\pi}{2}\right)}\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(x\ -\ \frac{\pi}{2}\right)\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^{-1}\left(\frac{1}{2}\right)\ =\ \frac{\pi}{6}\ \pm\ 2k\pi\ \ ] or *[tex \LARGE \ \ \frac{5\pi}{6}\ \pm\ 2k\pi]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{\pi}{2}\ =\ \frac{\pi}{6}\ \pm\ 2k\pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ =\ \frac{\pi}{6}\ +\ \frac{\pi}{2}\ \pm\ 2k\pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ =\ \frac{2\pi}{3}\ \pm\ 2k\pi]


which is in the given interval if and only if *[tex \Large k\ =\ 0]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ =\ \frac{2\pi}{3}]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{5\pi}{2}\ =\ \frac{\pi}{6}\ \pm\ 2k\pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ =\ \frac{5\pi}{6}\ +\ \frac{\pi}{2}\ \pm\ 2k\pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ =\ \frac{4\pi}{3}\ \pm\ 2k\pi]


which is not in the given interval for any values of *[tex \Large k]


Therefore the solution set contains the single element *[tex \Large \frac{2\pi}{3}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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