Question

The equation $\theta = 2n\pi \pm \left( \frac{\pi}{2} - 2\theta \right) \Rightarrow \theta \pm 2\theta = 2n\pi \pm \frac{\pi}{2}$is solvable for.

• A. $3\theta = 2n\pi + \frac{\pi}{2} \Rightarrow \theta = \frac{1}{3}\left( 2n\pi + \frac{\pi}{2} \right)$
• B. $- \theta = 2n\pi - \frac{\pi}{2} \Rightarrow \theta = - \left( 2n\pi - \frac{\pi}{2} \right)$
• C. $\theta$
• D. $\pi$

Answer 1

Answer: (C)

$\theta$

Answer 2

$\sin x + \sin y + \sin z = - 3$

⇒ $x = y = z = \frac{3\pi}{2},$

⇒ $x,y,z \in \lbrack 0,2\pi\rbrack.$

Let $\sin 2\theta = \cos\theta \Rightarrow \cos\theta = \cos\left( \frac{\pi}{2} - 2\theta \right)$. Then the given equation becomes

$\Rightarrow$,

where $\theta = 2n\pi \pm \left( \frac{\pi}{2} - 2\theta \right) \Rightarrow \theta \pm 2\theta = 2n\pi \pm \frac{\pi}{2}$, $3\theta = 2n\pi + \frac{\pi}{2} \Rightarrow \theta = \frac{1}{3}\left( 2n\pi + \frac{\pi}{2} \right)$

For real, discriminant $- \theta = 2n\pi - \frac{\pi}{2} \Rightarrow \theta = - \left( 2n\pi - \frac{\pi}{2} \right)\theta\pi$ $\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$ $30^{o},90^{o},150^{o}$

Also $2 - 2\cos^{2}\theta = 3\cos\theta$

$\Rightarrow$ $\cos\theta = \frac{- 3 \pm \sqrt{9 + 16}}{4} = \frac{- 3 \pm 5}{4}$. Thus $\cos\theta = \frac{1}{2} = \cos\left( \frac{\pi}{3} \right)$.