If $\cos {40^o} = x$ and $\cos \theta = 1 - 2{x^2}$, then the possible values of $\theta $ lying between ${0^o}$ and ${360^o}$is

The solution of the equation $cos^2\theta\, +\, sin\theta\, + 1\, =\, 0$ lies in the interval

Let $S=\{\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\}$.

Then $\sum_{\theta \in S } \sin ^2\left(\theta+\frac{\pi}{4}\right)$ is equal to

Number of solutions of $\sqrt {\tan \theta }  = 2\sin \theta ,\theta  \in \left[ {0,2\pi } \right]$ is equal to 

Let $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ and $\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)$, then $\frac{1}{6}(\beta-14)^2$ is equal to

The general solution of $\frac{{\tan \,2x\, - \,\tan \,x}}{{1\, + \,\tan \,x\,\tan \,2x}}\, = \,1$ is