Trigonometry in Algebra 2

Algebra Level 5

$\large \sqrt[3]{{\displaystyle\frac{{{x^9} + 9{x^2} - 1}}{3}}} = 2x + 1$

Let real numbers $x_1,x_2, x_3, \cdots x_n$ be the solutions of the equation above. Then $\displaystyle\frac{1}{\pi }\sum\limits_{i = 1}^n \arccos \frac{{{x_i}}}{2} = \frac{a}{b}$ , where $a$ and $b$ are positive coprime integers. Find $a+b$ .

Hint: Need to apply some calculus.

This problem is part of this set .

The answer is 22.

1 solution

Hugh Sir
Nov 2, 2018

$\sqrt[3]{\dfrac{x^{9}+9x^{2}-1}{3}} = 2x+1$

$x^{9}+9x^{2}-1 = 3(2x+1)^{3}$

$x^{9}+9x^{2}-1 = 24x^{3}+36x^{2}+18x+3$

$x^{9}-24x^{3}-27x^{2}-18x-4 = 0$

$(x^{3}-3x-1)(x^{6}+3x^{4}+x^{3}+9x^{2}+6x+4) = 0$

Since $x^{6}+3x^{4}+x^{3}+9x^{2}+6x+4 > 0$ for every real number $x$ , the equation to be considered is

$x^{3}-3x-1 = 0$

Substituting $x = 2\cos y$ into the equation yields

$8\cos^{3} y - 6\cos y - 1 = 0$

$4\cos^{3} y - 3\cos y = \dfrac{1}{2}$

$\cos 3y = \dfrac{1}{2}$

$3y = \dfrac{\pi}{3}, \dfrac{5\pi}{3}, \dfrac{7\pi}{3}$

$y = \dfrac{\pi}{9}, \dfrac{5\pi}{9}, \dfrac{7\pi}{9}$

Note that $y = \arccos \dfrac{x}{2}$ .

Then $\dfrac{1}{\pi} \sum \limits_{i=1}^3 \arccos \dfrac{x_i}{2} = \dfrac{1}{9} + \dfrac{5}{9} + \dfrac{7}{9} = \dfrac{13}{9} = \dfrac{a}{b}$ .

Therefore, $a+b = 13+9 = 22$ .

Trigonometry in Algebra 2

The answer is 22.

1 solution

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