Question

If sum of all the solutions of the equation $8cosx \cdot (cos\frac{\pi}{6} + x) \cdot cos(\frac{\pi}{6} - x) - \frac{1}{2} = 1$ in $(0, \pi)$ is $k\pi$, then k is equal to:

• A. 13/9
• B. 20/9
• C. 1
• D. 13/6

Answer 1

13/9

Answer 2

The given equation is: $8\cos x \cdot (\cos\frac{\pi}{6} + x) \cdot \cos(\frac{\pi}{6} - x) - \frac{1}{2} = 1$

Simplify the product of cosines: $(\cos\frac{\pi}{6} + x) \cdot \cos(\frac{\pi}{6} - x) = \frac{1}{2}[\cos((\frac{\pi}{6} + x) + (\frac{\pi}{6} - x)) + \cos((\frac{\pi}{6} + x) - (\frac{\pi}{6} - x))]$ $= \frac{1}{2}[\cos(\frac{\pi}{3}) + \cos(2x)]$ $= \frac{1}{2}[\frac{1}{2} + \cos(2x)]$

Substitute this back into the equation: $8\cos x \cdot \frac{1}{2}[\frac{1}{2} + \cos(2x)] - \frac{1}{2} = 1$ $4\cos x [\frac{1}{2} + \cos(2x)] = \frac{3}{2}$ $2\cos x + 4\cos x \cos(2x) = \frac{3}{2}$

Use the product-to-sum formula for $4\cos x \cos(2x)$: $4\cos x \cos(2x) = 2[2\cos(2x)\cos x] = 2[\cos(2x+x) + \cos(2x-x)] = 2[\cos(3x) + \cos x]$

Substitute this back into the equation: $2\cos x + 2(\cos(3x) + \cos x) = \frac{3}{2}$ $2\cos x + 2\cos(3x) + 2\cos x = \frac{3}{2}$ $4\cos x + 2\cos(3x) = \frac{3}{2}$ Multiply by 2: $8\cos x + 4\cos(3x) = 3$

Use the triple angle identity for cosine: $\cos(3x) = 4\cos^3 x - 3\cos x$. $8\cos x + 4(4\cos^3 x - 3\cos x) = 3$ $8\cos x + 16\cos^3 x - 12\cos x = 3$ $16\cos^3 x - 4\cos x - 3 = 0$

Let $y = \cos x$. The equation is $16y^3 - 4y - 3 = 0$. The roots of this equation are $y_1 = \cos(\pi/9)$, $y_2 = \cos(5\pi/9)$, $y_3 = \cos(7\pi/9)$. Since $x \in (0, \pi)$, $\cos x$ takes values in $(-1, 1)$. The values $\cos(\pi/9)$, $\cos(5\pi/9)$, and $\cos(7\pi/9)$ are all in $(-1, 1)$. The solutions for $x$ in $(0, \pi)$ are: $x_1 = \arccos(\cos(\pi/9)) = \pi/9$ $x_2 = \arccos(\cos(5\pi/9)) = 5\pi/9$ $x_3 = \arccos(\cos(7\pi/9)) = 7\pi/9$

The sum of these solutions is: $S = \frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} = \frac{1+5+7}{9}\pi = \frac{13\pi}{9}$

Given that the sum of solutions is $k\pi$, we have $k\pi = \frac{13\pi}{9}$. Therefore, $k = \frac{13}{9}$.