<s>[INST] If $ \cos(x) \cos(2x) \cos(3x) = \frac{4}{7} $ find $ \frac{1}{\cos^2{x}}+\frac{1}{\cos^2{2x}} + \frac{1}{\cos^2{3x}} $ 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 $
 [/INST] 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: $$S = 4.54217, S = 10.673$$
</s>