Question

If \(sin3α = 4sinα\cdot sin(x+α)\cdot sin(x-α)\) where \(α\neq nπ,n\in Z\), then all possible values of \(x\) are given as

• A. \(x = nπ\pm \frac{π}{3},n\in Z\)
• B. \(x = nπ\pm \frac{π}{4},n\in Z\)
• C. \(x = nπ\pm \frac{π}{6},n\in Z\)
• D. \(x = nπ\pm \frac{π}{2},n\in Z\)

Hint:

Use \(\sin(x+\alpha)\sin(x-\alpha) = \sin^2x - \sin^2\alpha\) and \(\sin3\alpha = 3\sin\alpha - 4\sin^3\alpha\).

Answer 1

The Correct Option is A

Step 1: Plan:
Compare with the known triple angle product formula $\sin 3\alpha = 4\sin\alpha\sin(60^{\circ}-\alpha)\sin(60^{\circ}+\alpha)$.

Step 2: Compare:
Here $\sin(60^{\circ}+\alpha)\sin(60^{\circ}-\alpha) = \sin^260^{\circ} - \sin^2\alpha$. In the given equation we have $\sin^2 x - \sin^2\alpha$ in the same place.
Since the equation holds with $\alpha$ arbitrary (not a multiple of $\pi$), we need $\sin^2x = \sin^260^{\circ} = \frac34$.

Step 3: General solution:
$\sin^2x = \sin^2\frac{\pi}{3}$ gives $x = n\pi \pm \frac{\pi}{3}$.

Final Answer:
All values of $x$ are $n\pi\pm\frac{\pi}{3}$, option (A). \[ \boxed{x = n\pi \pm \frac{\pi}{3},\ n\in Z} \]