Question

solve:
${\cot }^2\theta +3\cos ec\theta +3=0$

Answer 1

${\cot }^2\theta +3cosec\theta +3=0$

$\frac{{\cos }^2\theta }{{\sin }^2\theta }+\frac{3}{\sin \theta }+3=0$

$\frac{{\cos }^2\theta +3\sin \theta +3{\sin }^2\theta }{{\sin }^2\theta }=0$

${\cos }^2\theta +3{\sin }^2\theta +3\sin \theta =0$

$1+2{\sin }^2\theta +3\sin \theta =0$

put $\sin \theta =x$ in above equation we get

$2x^2+3x+1=0$

solution of above polynomial is given by:

$x=\frac{-3\pm \sqrt{3^2-4\times 2\times 1}}{2^2}$

$x=\frac{-3\pm \sqrt{9-8}}{4}$

$x=\frac{-3\pm \sqrt{1}}{4}$

$x=\frac{-3\pm 1}{4}$

Hence,

$x=-1$ or $x=-\frac{1}{2}$

Hence,

$\sin \theta =-1$ or $\sin \theta =-\frac{1}{2}$

$\to \theta =\frac{(4n+3)\pi }{2}$ or $\theta =\frac{(12n+7)\pi }{6}=\frac{(12n+11)\pi }{6}$

where $n=0,1,2,3,.....$