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If $\cos(x) \cos(2x) \cos(3x) = \dfrac{4}{7} $ and $S=\dfrac{1}{\cos^2{x}}+\dfrac{1}{\cos^2{2x}} + \dfrac{1}{\cos^2{3x}} $ when $ S \in \mathbb{R}^{+}$ then $ S = ? $

P.S. I have tried that , but failed many times. Because I suppose that $\cos(x) ,\cos(2x) , \cos(3x)$ be root of $8\cos^{6}(x) -10\cos^{4}(x) + 3\cos^{2}(x) - \frac{4}{7} = 0 $

wythagoras
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ABCDEFG user157844
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  • What is the source of the problem? – lab bhattacharjee Dec 22 '15 at 06:23
  • The problem really seems to be too interesting, unfortunately does not seem to be complete – lab bhattacharjee Dec 22 '15 at 06:25
  • $$8=7\cos2x(2\cos x\cos3x)=7\cos2x(\cos2x+\cos4x)=7\cos2x(\cos2x+2\cos^22x-1)$$

    $$\iff14\cos^32x+7\cos^22x-7\cos2x-8=0$$

     – lab bhattacharjee Dec 22 '15 at 06:39
  • oh , thank you. But if I change $\cos(x) \cos(2x) \cos(3x) = \frac{4}{7} $ to $8cos^{6}(x) -10cos^{4}(x) + 3cos^{2}(x) - \frac{4}{7} = 0 $ and then solve as cubic equation . is it possible to solve that ? – ABCDEFG user157844 Dec 22 '15 at 06:46
  • Wolframalpha indicates the answer is no (well if you use the general solution formula then yes but I would not say that is computable by human hand). However this does not reject the existence of a possible smart way to directly derive an answer to $S$ without solving the value of $\cos(x)$. – cr001 Dec 22 '15 at 07:15
  • @user157844 The solution to any cubic equation can be found here: http://www.math.vanderbilt.edu/~schectex/courses/cubic/ – Simply Beautiful Art Dec 22 '15 at 15:08

1 Answers1

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Question: $\cos(x) \cos(2x) \cos(3x) = \frac {4}{7}$ and $S=\dfrac{1}{\cos^2{x}}+\dfrac{1}{\cos^2{2x}} + \dfrac{1}{\cos^2{3x}} $ when $ S \in \mathbb{R}^{+}$ then $ S = ?$

Working Out: let $a = \cos(x)$, $b = \cos(2x)$, $c = \cos(3x)$, then: $$abc = \frac {4}{7}$$ $$S=\dfrac{1}{a^2}+\dfrac{1}{b^2} + \dfrac{1}{c^2} $$

Because $\cos(2x) = 2\cos^2(x) - 1$, $\therefore b = 2a^2 -1$.

Because $\cos(3x) = \cos(x)\cos(2x) - \sin(x)(2\sin(x)\cos(x)) = ab - 2a\sin^2(x)$.

As $2a\sin^2(x) = 2a(1- \cos^2(x)) = 2a(1-a^2)$

$\therefore \cos(3x) = ab(2a^2-1) - 2a(1-a^2) = 4a^3 - 3a$

$\therefore c = 4a^3 - 3a$

So we get the following equation: $$a(2a^2 - 1)(4a^3 - 3a) = \frac {4}{7}$$ $$56a^6 - 70a^4 + 21a^2 - 4 = 0$$

As I have know special way of solving such polynomials I turned to wolfram alpha and got an approximate root of $a = -0.96392$ and $a = 0.96392$. But $a = \cos(x)$, so when $a = -0.96292$ $x = 164.562(3.dp)$ and when $a = 0.96392$ $x = 15.4378 (4.dp)$. Putting those values into the equation of $S$, we get:

$$S=\dfrac{1}{\cos^2(164.562)}+\dfrac{1}{\cos^2(2 * 164.562)} + \dfrac{1}{\cos^2(3 * 164.562)} $$

$$\therefore S = 10.673$$

Or

$$S=\dfrac{1}{\cos^2(15.4378)}+\dfrac{1}{\cos^2(2 * 15.4378)} + \dfrac{1}{\cos^2(3 * 15.4378)} $$

$$\therefore S = 4.54217$$

Answer: $$S = 4.54217, S = 10.673$$

frog1944
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