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

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#### anemone

##### MHB POTW Director
Staff member
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

$8^{1+|\cos x|+\cos^2 x+|\cos^3 x|+\cdots+\infty}=4^3$

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solutions:

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

. . . . . . . $$1 + |\cos x| + |\cos^2\!x| + |\cos^3\!x| + \cdots \:=\:2$$

We 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: .$$x \;=\;\pm\tfrac{\pi}{3},\;\pm\tfrac{2\pi}{3}$$

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