Question

If $\sin^{2}x\lbrack 4\sin^{2}x - 1 - (1 - \sin^{2}x)\rbrack = 0$, then the value of $\Rightarrow$is.

• A. $\sin^{2}x\lbrack 5\sin^{2}x - 2\rbrack = 0$
• B. $\Rightarrow$
• C. $\sin x = 0$
• D. None of these

Answer 1

Answer: (B)

$\Rightarrow$

Answer 2

$\theta = \frac{1}{2}\left\lbrack n\pi + ( - 1)^{n}\sin^{- 1}\left( \frac{3}{4} \right) \right\rbrack$

$\cos p\theta = \cos q\theta \Rightarrow p\theta = 2n\pi \pm q\theta \Rightarrow \theta = \frac{2n\pi}{p \pm q}4 + 2\sin^{2}x = 5$

$\Rightarrow$ $\sin^{2}x = \frac{1}{2} = \sin^{2}\frac{\pi}{4} \Rightarrow x = n\pi \pm \frac{\pi}{4}$ $3\sin\alpha - 4\sin^{3}\alpha = 4\sin\alpha(\sin^{2}x - \sin^{2}\alpha)$ $\therefore$.

Trick : Since $\sin^{2}x = \left( \frac{\sqrt{3}}{2} \right)^{2}$ satisfies the equation and therefore the general value should be $\sin^{2}x = \sin^{2}\pi/3$.