Question:

Step 1: Simplify the product of cosines using the product-to-sum identity. We need to simplify $\cos\left(\frac{\pi}{3} - x\right)\cos\left(\frac{\pi}{3} + x\right)$ using the identity: $$\cos(A-B)\cos(A+B) = \cos^2 A - \sin^2 B$$ With $A = \frac{\pi}{3}$ and $B = x$: $$\cos\left(\frac{\pi}{3} - x\right)\cos\left(\frac{\pi}{3} + x\right) = \cos^2\left(\frac{\pi}{3}\right) - \sin^2 x = \frac{1}{4} - \sin^2 x$$ Step 2: Apply the triple angle identity for the complete expression. There is a useful identity for the product of three cosines: $$\cos\theta \cdot \cos\left(\frac{\pi}{3} - \theta\right) \cdot \cos\left(\frac{\pi}{3} + \theta\right) = \frac{1}{4}\cos 3\theta$$ Applying this with $\theta = x$: $$\cos x \cos\left(\frac{\pi}{3} - x\right)\cos\left(\frac{\pi}{3} + x\right) = \frac{1}{4}\cos 3x$$ Step 3: Substitute into the original equation and solve for $x$. The original equation becomes: $$\frac{1}{4}\cos 3x = \frac{1}{4}$$ Multiplying both sides by 4: $$\cos 3x = 1$$ Step 4: Find all values of $3x$ that satisfy the equation. The general solution for $\cos 3x = 1$ is: $$3x = 2n\pi, \quad n \in \mathbb{Z}$$ Therefore: $$x = \frac{2n\pi}{3}$$ Step 5: Identify solutions in the interval $(0, 2\pi)$. For $x \in (0, 2\pi)$, we need: $$0 < \frac{2n\pi}{3} < 2\pi$$ $$0 < n < 3$$ So $n \in \{1, 2\}$, giving us: - When $n = 1$: $x = \frac{2\pi}{3}$ - When $n = 2$: $x = \frac{4\pi}{3}$ Step 6: Calculate the sum of all solutions. $$\text{Sum} = \frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$$ The sum of solutions in $(0, 2\pi)$ is $\boxed{2\pi}$, which corresponds to **Option 3**.