Question

Solve :

${(1+2\sin \theta )}^2+{(\sqrt{3}\tan \theta -1)}^2=0$

Answer 1

${(1+2\sin \theta )}^2+{(\sqrt{3}\tan \theta -1)}^2=0$

this is only possible only if both the terms are zero

$\Rightarrow 1+2\sin \theta =0$ and $\sqrt{3}\tan \theta -1=0$

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

$\Rightarrow \theta$ lies in third quadrant hence $\theta =\frac{7\pi }{6}$

Hence general solution is, $\theta =2n\pi +\frac{7\pi }{6}$