MHB Solving for x: Interval -π to π | MHB POTW #97 (2/3/2014)

[anemone]

Hello MHB,

I am truly honored and delighted to have been asked to take over posting the POTW for Secondary School/High School Students. This week's problem is as follows:

Find all the values of $x$ lying in the interval $(-\pi,\,\pi)$ which satisfy the equation

$8^{1+|\cos x|+\cos^2 x+|\cos^3 x|+\cdots+\infty}=4^3$--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. kaliprasad
2. soroban
3. MarkFL
4. lfdahl
5. Pranav

Honorable mention goes to both magneto and mente oscura for finding two of the correct $x$ values but missing the other two $x$ values.

soroban's solution:

Spoiler

We have: .(2^3)^{(1+|\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots)} \:=\: (2^2)^3

. . . . . . . . 2^{3(1+|\cos x| + |\cos^2\!x + |\cos^3\!x| + \cdots)} \:=\:2^6

Hence: .3(1+|\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots) \;=\;6

. . . . . . . 1 + |\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots \:=\:2We have an infinite series with first term a = 1, common ratio r = |\cos x|

Its sum is: .\frac{1}{1-|\cos x|}We have: .\frac{1}{1-|\cos x|} \:=\:2 \quad\Rightarrow\quad 1-|\cos x| \:=\:\tfrac{1}{2}

. . . . . . . . |\cos x| \:=\:\tfrac{1}{2}Therefore: .x \;=\;\pm\tfrac{\pi}{3},\;\pm\tfrac{2\pi}{3}