Let $$f\left( x \right) = x\sin \,\pi x,\,x > 0.$$ Then for all natural numbers $$n,\,f'\left( x \right)$$ vanishes at

Let $$\theta ,\,\varphi \, \in \,\left[ {0,2\pi } \right]$$ be such that
$$2\cos \theta \left( {1 - \sin \,\varphi } \right) = {\sin ^2}\theta \,\,\left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)\cos \varphi - 1,\,\tan \left( {2\pi - \theta } \right) > 0$$ and $$ - 1 < \sin \theta \, < - {{\sqrt 3 } \over 2},$$

then $$\varphi $$ cannot satisfy

For $$0 < \theta < {\pi \over 2},$$ the solution (s) of $$$\sum\limits_{m = 1}^6 {\cos ec\,\left( {\theta + {{\left( {m - 1} \right)\pi } \over 4}} \right)\,\cos ec\,\left( {\theta + {{m\pi } \over 4}} \right) = 4\sqrt 2 } $$$ is (are)

If $${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5},$$ then