The number of real solutions to the equation \(\left | sec(x)-2\right |=k\), where \(-2\pi\le x \le2\pi\)

is determined by the value of the constant k.

Find the number of real solutions for all values of k, given that \(k\epsilon R\).


Here is my attempt:

\(\frac{1}{cos(x)}-2=\pm k\) for \(-2\pi\le x\le 2\pi\)

\(cos(x)=\frac{1}{2\pm k}\)

\(x=cos^{-1}(\frac{1}{2\pm k}) \) ( \(-2\pi\le x\le 2\pi\) ) 


\(-2\pi\le cos^{-1}(\frac{1}{2\pm k})\le 2\pi\) implying: \(1\le\frac{1}{2\pm k}\le1\).

Therefore, \(\frac{1}{2\pm k}=1\), so \(k=\pm1\)

Now, i subsituited k=1 and k=-1 in the original equation \(\left | sec(x)-2\right |=k\) and considering the domain, \(-2\pi\le x \le2\pi\)

found the solutions seperately for each case (I.e. k=1 and k=-1) Then counted them and they are 7 distinct solutions.

Is this correct? As I do not have a model answer to check. 

Thanks in advance.

 Jan 27, 2022

I just found a solution, what I did above is not correct.


-->Sketch the graph of \(y=\left |sec(x)-2 \right |\), and then consider y=k as k varies over R.

That is,

if k<0 -->No real solutions

if k=0 -->4 real solutions

if 0 8 real solutions

if k=1 -->7 real solutions

if 1 4 real solutions

if k=3 -->6 real solutions

if k>3 ---> 8 real solutions.

 Jan 27, 2022

here is the graphical solution




k less than 0                      0 solutions

k=0                                   4 solutions

k is between 0 and 1         8 solutions

k=1                                     7 solutions

k is between 1 and 3          4 solutions

k=3                                  6 solutions

kis bigger than 3               8 solutions

 Jan 27, 2022
edited by Melody  Jan 27, 2022

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