Question

If $\beta = \theta - \alpha$, then $y^{2} - 2y - 2(1 + \alpha) = 0$ (where $- 1 \leq y \leq 1$).

• A. $(\because\text{ } - 1 \leq \sin 2x \leq 1)$
• B. $\geq 0$
• C. $\Rightarrow$
• D. None of these

Answer 1

Answer: (A)

$(\because\text{ } - 1 \leq \sin 2x \leq 1)$

Answer 2

We have $\frac{\theta}{4} = k\pi$

$\theta = 2k\pi\theta = 4k\pi$

$\Rightarrow$ $\frac{\tan 3\theta - 1}{\tan 3\theta + 1} = \sqrt{3}$

$\Rightarrow$ $\tan\left( 3\theta - \frac{\pi}{4} \right) = \tan\frac{\pi}{3}$

$3\theta - (\pi/4) = n\pi + (\pi/3)3\theta = n\pi + \frac{7\pi}{12}$ and $\theta = \frac{n\pi}{3} + \frac{7\pi}{36}$

$2 - 2\sin^{2}x + 3\sin x - 3 = 0 \Rightarrow$and $(2\sin x - 1)(\sin x - 1) = 0$=0, $\Rightarrow$

Therefore, the general solution of $\sin x = \frac{1}{2}$ and $\sin x = 1$is$\Rightarrow$where $x = \frac{\pi}{6},\frac{5\pi}{6}$.