+0

+1
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Question:
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 $$k\epsilon R$$.

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Here is my attempt:

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

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

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

So:

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

Jan 27, 2022

#1
+1

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
#2
+117849
+1

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