Question

If $4S^{2} + 9S - 9 = 0$ then the general value $\therefore$is.

• A. $(S + 3)(4S - 3) = 0$
• B. $S = \frac{3}{4}$
• C. $S \neq - 3$
• D. $\sin 2\theta = \frac{3}{4} = \sin\alpha$

Answer 1

Answer: (C)

$S \neq - 3$

Answer 2

$\frac{\tan 2\theta + \tan 3\theta}{1 - \tan 2\theta\tan 3\theta} = \frac{1}{\sqrt{3}}$

$\tan 5\theta = \tan\frac{\pi}{6} \Rightarrow$

Dividing by $5\theta = n\pi + \frac{\pi}{6} \Rightarrow \theta = \left( n + \frac{1}{6} \right)\frac{\pi}{5}$ on both sides, we get

$\tan 2\theta = \cot\theta$

$\tan 2\theta = \tan\ \left( \frac{\pi}{2} - \theta \right) \Rightarrow$

$\Rightarrow$ $\frac{1}{\sin\theta} = 1 + \frac{\cos\theta}{\sin\theta} \Rightarrow \sin\theta + \cos\theta = 1$.