Question

If $\gamma \sin \theta =3,\gamma =4(1+\sin \theta ),0\le \theta \le 2\pi$, then $\theta =----$

• A. $\frac{\pi }{6},\frac{5\pi }{6}$
• B. $\frac{\pi }{3},\frac{2\pi }{3}$
• C. $\frac{\pi }{4},\frac{5\pi }{4}$
• D. $\frac{\pi }{2},\pi$

Answer 1

Answer: (A)

$\frac{\pi }{6},\frac{5\pi }{6}$

we get
$4(1+sin\theta )sin\theta =3$

$4{\sin }^2\theta +4\sin \theta -3=-0$
this gives $sin\theta =\frac{1}{2}or-\frac{3}{2}\left(rejected\right)$
so our solution set is $\theta =\frac{\pi }{6},\frac{5\pi }{6}$