The number of solutions of the equation $\tan \mathrm{\theta }+\cot 2\mathrm{\theta }=1$ lying in the interval (-π,π) is

$\tan \mathrm{\theta }+\cot 2\mathrm{\theta }=1\Rightarrow \tan \mathrm{\theta }+\frac{1-{\tan }^2\mathrm{\theta }}{2\tan \mathrm{\theta }}=1\Rightarrow 2{\tan }^2\mathrm{\theta }+1-{\tan }^2\mathrm{\theta }=2\tan \mathrm{\theta }$
$\Rightarrow \tan \mathrm{\theta }-2\tan \mathrm{\theta }+1=0\Rightarrow (\tan \mathrm{\theta }-1{)}^2=0\Rightarrow \tan \mathrm{\theta }=1\Rightarrow \mathrm{\theta }=\mathrm{\pi }/4,-\mathrm{\pi }/4\text{ in }(-\mathrm{\pi },\mathrm{\pi })$
$\therefore$ Number of solutions = 2.