Question

If $\sin^{2}x = \frac{1}{2} = \sin^{2}\frac{\pi}{4} \Rightarrow x = n\pi \pm \frac{\pi}{4}$, then the general values of $3\sin\alpha - 4\sin^{3}\alpha = 4\sin\alpha(\sin^{2}x - \sin^{2}\alpha)$ are.

• A. $\therefore$
• B. $\sin^{2}x = \left( \frac{\sqrt{3}}{2} \right)^{2}$
• C. $\sin^{2}x = \sin^{2}\frac{\pi}{3}$
• D. $x = n\pi \pm \frac{\pi}{3}$

Answer 1

Answer: (B)

$\sin^{2}x = \left( \frac{\sqrt{3}}{2} \right)^{2}$

Answer 2

$\theta = 2n\pi + \frac{\pi}{2}$; $\theta = 2n\pi$

i.e., $\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = 3$

or $\frac{1 - (1 - 2\sin^{2}\theta)}{1 + (2\cos^{2}\theta - 1)} = 3$.