If 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 $$[0, \pi]$$ is $$k\pi$$, then k is equal to:

We start with the given equation

$$8\cos x\;\Bigl(\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)-\tfrac12\Bigr)=1,\qquad x\in[0,\pi].$$

First we simplify the expression inside the big parentheses. We recall the product-to-sum identity

$$\cos A\cos B=\tfrac12\bigl[\cos(A+B)+\cos(A-B)\bigr].$$

Here we have $$A=\tfrac{\pi}{6}+x,\;B=\tfrac{\pi}{6}-x.$$ Substituting, we obtain

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right) =\tfrac12\Bigl[\cos\!\bigl((\tfrac{\pi}{6}+x)+(\tfrac{\pi}{6}-x)\bigr)+\cos\!\bigl((\tfrac{\pi}{6}+x)-(\tfrac{\pi}{6}-x)\bigr)\Bigr].$$

Simplifying the angles, $$(\tfrac{\pi}{6}+x)+(\tfrac{\pi}{6}-x)=\tfrac{\pi}{3}$$ and $$(\tfrac{\pi}{6}+x)-(\tfrac{\pi}{6}-x)=2x$$, so

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right) =\tfrac12\,[\cos(\tfrac{\pi}{3})+\cos(2x)].$$

We know $$\cos(\tfrac{\pi}{3})=\tfrac12,$$ hence

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)=\tfrac12\!\left[\tfrac12+\cos(2x)\right]=\tfrac14+\tfrac12\cos(2x).$$

Now subtract the $$\tfrac12$$ that appears in the original bracket:

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)-\tfrac12 =\Bigl(\tfrac14+\tfrac12\cos(2x)\Bigr)-\tfrac12 =-\tfrac14+\tfrac12\cos(2x).$$

Substituting this result back into the main equation, we get

$$8\cos x\;\Bigl(-\tfrac14+\tfrac12\cos(2x)\Bigr)=1.$$

Multiplying by $$8\cos x$$ term by term gives

$$8\cos x\left(-\tfrac14\right)+8\cos x\left(\tfrac12\cos(2x)\right)=1,$$

or

$$-2\cos x+4\cos x\cos(2x)=1.$$

We next express $$\cos(2x)$$ in terms of $$\cos x$$ using the double-angle formula

$$\cos(2x)=2\cos^2x-1.$$

Substituting this, we find

$$4\cos x\cos(2x)=4\cos x\bigl(2\cos^2x-1\bigr)=8\cos^3x-4\cos x.$$

Therefore

$$-2\cos x+\bigl(8\cos^3x-4\cos x\bigr)=1,$$ or $$8\cos^3x-6\cos x=1.$$

We notice the expression $$4\cos^3x-3\cos x$$, which is famous from the triple-angle identity

$$\cos(3x)=4\cos^3x-3\cos x.$$

Indeed,

$$8\cos^3x-6\cos x=2(4\cos^3x-3\cos x)=2\cos(3x).$$

So the equation reduces neatly to

$$2\cos(3x)=1\;\;\Longrightarrow\;\;\cos(3x)=\tfrac12.$$

Let us set $$y=3x.$$ Because $$x\in[0,\pi],$$ we have $$y\in[0,3\pi].$$ We now solve

$$\cos y=\tfrac12,\qquad y\in[0,3\pi].$$

The general solutions of $$\cos y=\tfrac12$$ are

$$y=2n\pi\pm\tfrac{\pi}{3},\qquad n\in\mathbb Z.$$

We list the values lying in $$[0,3\pi]$$:

• For $$n=0$$: $$y=\tfrac{\pi}{3}$$ (while $$-\tfrac{\pi}{3}$$ is negative and hence outside the interval).
• For $$n=1$$: $$y=2\pi-\tfrac{\pi}{3}=\tfrac{5\pi}{3}$$ and $$y=2\pi+\tfrac{\pi}{3}=\tfrac{7\pi}{3}$$; both are within $$[0,3\pi]$$.
• $$n=2$$ would give values $$\gt \,3\pi$$, so we stop here.

Thus $$y\in\Bigl\{\tfrac{\pi}{3},\;\tfrac{5\pi}{3},\;\tfrac{7\pi}{3}\Bigr\}.$$

We now convert back to $$x$$ using $$x=\tfrac{y}{3}$$:

$$x_1=\tfrac{\pi}{9},\qquad x_2=\tfrac{5\pi}{9},\qquad x_3=\tfrac{7\pi}{9}.$$

All three lie in the required interval $$[0,\pi]$$, so they are the complete set of solutions.

Their sum is

$$x_1+x_2+x_3=\tfrac{\pi}{9}+\tfrac{5\pi}{9}+\tfrac{7\pi}{9} =\tfrac{(1+5+7)\pi}{9}=\tfrac{13\pi}{9}.$$

Comparing with the form $$k\pi$$, we see $$k=\tfrac{13}{9}.$$

Hence, the correct answer is Option C.