Question

General solution of the equation $\sin \theta \sin ({60}^{\circ }+\theta )\sin ({60}^{\circ }-\theta )=\frac{\sqrt{5}-1}{16}$ is

• A. $n\pi +(-1{)}^n\frac{\pi }{10},n\in Z$
• B. $\frac{n\pi }{3}+(-1{)}^n\frac{\pi }{10},n\in Z$
• C. $\frac{n\pi }{3}+(-1{)}^n\frac{\pi }{30},n\in Z$
• D. $n\pi +(-1{)}^n\frac{\pi }{30},n\in Z$

Answer 1

The correct option is C $\frac{n\pi }{3}+(-1{)}^n\frac{\pi }{30},n\in Z$

$\sin \theta \sin ({60}^{\circ }+\theta )\sin ({60}^{\circ }-\theta )=\frac{\sqrt{5}-1}{16}$
$\Rightarrow \frac{1}{4}\sin 3\theta =\frac{\sqrt{5}-1}{16}$
$\Rightarrow \sin 3\theta =\frac{\sqrt{5}-1}{4}=\sin {18}^{\circ }$
$\Rightarrow \sin 3\theta =\sin \frac{\pi }{10}$
So, the general solution of the above equation is
$3\theta =n\pi +(-1{)}^n\frac{\pi }{10},n\in Z$
$\Rightarrow \theta =\frac{n\pi }{3}+(-1{)}^n\frac{\pi }{30},n\in Z$