Welcome to our community

Be a part of something great, join today!

Problem of the week #97 - February 3rd, 2014

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
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 POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
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.:eek:

soroban's solution:
We have: .[tex](2^3)^{(1+|\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots)} \:=\: (2^2)^3[/tex]

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

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

. . . . . . . [tex]1 + |\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots \:=\:2[/tex]


We have an infinite series with first term [tex]a = 1[/tex], common ratio [tex]r = |\cos x|[/tex]

Its sum is: .[tex]\frac{1}{1-|\cos x|} [/tex]


We have: .[tex]\frac{1}{1-|\cos x|} \:=\:2 \quad\Rightarrow\quad 1-|\cos x| \:=\:\tfrac{1}{2}[/tex]

. . . . . . . . [tex]|\cos x| \:=\:\tfrac{1}{2}[/tex]


Therefore: .[tex]x \;=\;\pm\tfrac{\pi}{3},\;\pm\tfrac{2\pi}{3}[/tex]
 
Status
Not open for further replies.