Question

Values of $\cos 2x = 0, - 1$ satisfying $\therefore$ are .

• A. $2x = \left( n + \frac{1}{2} \right)\ \pi$
• B. $(2n + 1)\ \pi$
• C. $\Rightarrow$
• D. $x = (2n + 1)\frac{\pi}{4}\$

Answer 1

Answer: (D)

$x = (2n + 1)\frac{\pi}{4}\$

Answer 2

$2\sin^{2}x + \sin^{2}2x = 2$

$\Rightarrow$ $\sin^{2}2x = 2\cos^{2}x2\cos^{2}x\cos 2x = 0$ $x = (2n + 1)\frac{\pi}{2}\text{ or }x = (2n + 1)\frac{\pi}{4}$ or $\therefore$.