Question

If $x = \frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\frac{5\pi}{6},\frac{5\pi}{4},\frac{7\pi}{6},\frac{7\pi}{4},\frac{9\pi}{6},\frac{11\pi}{6}$then x =

• A. $\sin^{4}x + \cos^{4}x + \sin 2x + \alpha = 0$
• B. $(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x + \sin 2x + \alpha = 0$
• C. $\sin^{2}2x - 2\sin 2x - 2 - 2\alpha = 0$
• D. None of these

Answer 1

Answer: (A)

$\sin^{4}x + \cos^{4}x + \sin 2x + \alpha = 0$

Answer 2

The given equation can be put in the form

$3\theta = n\pi + \frac{\pi}{3}$

$\theta = (3n + 1)\frac{\pi}{9}1 - \cos\theta = \sin\theta.\sin\frac{\theta}{2}$

$2\sin^{2}\frac{\theta}{2} = 2\sin\frac{\theta}{2}.\cos\frac{\theta}{2}.\sin\frac{\theta}{2}$ $\sin ^ { 2 } x \left[ 5 \sin ^ { 2 } x - 2 \right] = 0$ $\sin\frac{\theta}{2} = 02\sin^{2}\frac{\theta}{4} = 0$or $\sin\frac{\theta}{2}$.

Hence $\sin\frac{\theta}{4} = 0$ is the required answer.