The number of solutions of the pair of equations $ 2 \sin ^2 \theta-\cos 2 \theta=0 $, $ 2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is

The number of elements in the set $S=\left\{x \in R : 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$ is$.....$

Let $P = \left\{ {\theta :\sin \,\theta  - \cos \,\theta  = \sqrt 2 \,\cos \,\theta } \right\}$ and $Q = \left\{ {\theta :\sin \,\theta  + \cos \,\theta  = \sqrt {2\,} \sin \,\theta } \right\}$ be two sets. Then

The value of expression $\frac{{2(\sin {1^o} + \sin {2^o} + \sin {3^o} + ..... + \sin {{89}^o})}}{{2(\cos {1^o} + \cos {2^o} + .... + \cos {{44}^o}) + 1}}$ equals

If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ then 

$\sum\limits_{r = 1}^{100} {\frac{{\tan \,{2^{r - 1}}}}{{\cos \,{2^r}}}} $ is equal to