If the sum of all the solutions of the equati | vedclass

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(A) Given equation: $8 \cos x \left\{ \cos \left( \frac{\pi}{6} + x \right) \cos \left( \frac{\pi}{6} - x \right) - \frac{1}{2} \right\} = 1$ Using th

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$\frac{13}{9}$

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(A) Given equation: $8 \cos x \left\{ \cos \left( \frac{\pi}{6} + x \right) \cos \left( \frac{\pi}{6} - x \right) - \frac{1}{2} \right\} = 1$ Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $4 \cos x \left\{ \cos \left( \frac{\pi}{3} \right) + \cos(2x) - 1 \right\} = 1$ Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$: $4 \cos x \left\{ \frac{1}{2} + \cos 2x - 1 \right\} = 1$ $4 \cos x \left\{ \cos 2x - \frac{1}{2} \right\} = 1$ $4 \cos x \left\{ 2 \cos^2 x - 1 - \frac{1}{2} \right\} = 1$ $8 \cos^3 x - 6 \cos x = 1$ $2(4 \cos^3 x - 3 \cos x) = 1$ $2 \cos 3x = 1 \Rightarrow \cos 3x = \frac{1}{2}$ For $x \in [0, \pi]$,$3x \in [0, 3\pi]$. Solutions for $3x$: $3x = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3} = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$. Thus,$x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$. Sum of solutions $= \frac{\pi + 5\pi + 7\pi}{9} = \frac{13\pi}{9}$. Given sum $= k\pi$,so $k = \frac{13}{9}$.

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

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