Question:

Step 1: State the given equation and a relevant identity. The given equation is: $$ \sin 3a = 4\sin a \sin(x+a)\sin(x-a) $$ A known triple angle identity for sine in product form is: $$ \sin 3a = 4\sin a \sin\left(\frac{\pi}{3} - a\right)\sin\left(\frac{\pi}{3} + a\right) $$ Step 2: Compare the expressions. Comparing the given equation with the identity, we must have: $$ 4\sin a \sin(x+a)\sin(x-a) = 4\sin a \sin\left(\frac{\pi}{3} - a\right)\sin\left(\frac{\pi}{3} + a\right) $$ Assuming $\sin a \neq 0$, we can divide both sides by $4\sin a$: $$ \sin(x+a)\sin(x-a) = \sin\left(\frac{\pi}{3} - a\right)\sin\left(\frac{\pi}{3} + a\right) $$ This equality holds if the arguments are equivalent, specifically: $$ x+a = \frac{\pi}{3} + a \quad \text{and} \quad x-a = \frac{\pi}{3} - a $$ Step 3: Determine the value of x. From the first equality, $x+a = \frac{\pi}{3} + a$, subtracting $a$ from both sides yields: $$ x = \frac{\pi}{3} $$ From the second equality, $x-a = \frac{\pi}{3} - a$, adding $a$ to both sides yields: $$ x = \frac{\pi}{3} $$ Step 4: State the general solution. Considering the periodicity and symmetry inherent in the product of sine functions, the general solution for $x$ is: $$ x = n\pi \pm \frac{\pi}{3}, \quad \text{where } n \in \mathbb{Z} $$