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: Understanding the Question:
The equation has to hold for the given \(\alpha\), with \(\sin\alpha \ne 0\). We can divide by \(\sin\alpha\).

Step 2: Key identities:
\(\sin(x+\alpha)\sin(x-\alpha) = \sin^2x - \sin^2\alpha\).
\(\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha\).

Step 3: Solve:
Substitute: \(3\sin\alpha - 4\sin^3\alpha = 4\sin\alpha(\sin^2x - \sin^2\alpha)\).
Divide by \(\sin\alpha\): \(3 - 4\sin^2\alpha = 4\sin^2x - 4\sin^2\alpha\).
The \(\sin^2\alpha\) terms cancel, so \(4\sin^2 x = 3\), i.e. \(\sin^2 x = \frac{3}{4}\).
\[ \sin x = \pm\frac{\sqrt3}{2} \Rightarrow x = n\pi \pm \frac{\pi}{3},\ n\in Z \]

Step 4: Other options:
\(\frac{\pi}{4}\) gives \(\sin^2x = \frac12\), \(\frac{\pi}{6}\) gives \(\frac14\), and \(\frac{\pi}{2}\) gives \(1\). None equals \(\frac34\).

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