Question

The roots of the equation $\theta - \alpha = 0^{ο},180^{o} \Rightarrow \theta = \alpha,\ 180^{o} + \alpha$is.

• A. $3\sin^{2}x - 7\sin x + 2 = 0$
• B. $\Rightarrow$
• C. $3\sin^{2}x - 6\sin x - \sin x + 2 = 0$
• D. None of these

Answer 1

Answer: (B)

$\Rightarrow$

Answer 2

We have, $\sec^{2}\theta - \tan^{2}\theta = 1$

⇒ $\Rightarrow$

⇒ $\sec\theta - \tan\theta = \frac{1}{\sqrt{3}}$⇒ $\tan\theta = \frac{1}{2}\left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = \frac{1}{\sqrt{3}} = \tan\left( \frac{\pi}{6} \right)$or $\Rightarrow$

⇒ $\theta = n\pi + \frac{\pi}{6}$ = 0 or $\therefore$⇒ $0 \leq \theta \leq 2\pi$ or $\frac{\pi}{6}$.

Hence, $\frac{7\pi}{6}$ or $\sin 5x + \sin 3x + \sin x = 0$, $\Rightarrow$.