Question

The value(s) of $\theta$ lying between $0\&2\pi$ satisfying the equation $r\sin \theta =\sqrt{3}\&r+4\sin \theta =2(\sqrt{3}+1)$ is/are

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

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{\pi }{6}$
B $\frac{\pi }{3}$
C $\frac{2\pi }{3}$
D $\frac{5\pi }{6}$

$r\sin \theta =\sqrt{3}\Rightarrow r=\frac{\sqrt{3}}{\sin \theta }$

Then $\frac{\sqrt{3}}{\sin \theta }+4\sin \theta =2(\sqrt{3}+1)$

$\Rightarrow \sqrt{3}+4{\sin }^2\theta =2(\sqrt{3}+1)\sin \theta$

$\Rightarrow 4{\sin }^2\theta -2\sqrt{3}\sin \theta -2sin\theta +\sqrt{3}$

$\Rightarrow (2\sin \theta -\sqrt{3})(2\sin \theta -1)=0$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2},\sin \theta =\frac{1}{2}$

in $(0,2\pi )$

Therefore $\Rightarrow \theta =\frac{\pi }{6},\frac{\pi }{3},\frac{2\pi }{3},\frac{5\pi }{6}$