If $\mathrm{n}$ is the number of solutions of the equation
$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$
and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :
The only value of $x$ for which ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$ holds, is
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
The solution set of the system of equation
$x\,\, + \,\,y\,\, = \,\,\frac{{2\pi }}{3},\,{\rm{cos}}\,{\rm{x + }}\,{\rm{ cos}}\,{\rm{y}}\,{\rm{ = }}\,\frac{3}{2},$ where $x$ and $y$ are real in
If equation in variable $\theta, 3 tan(\theta -\alpha) = tan(\theta + \alpha)$, (where $\alpha$ is constant) has no real solution, then $\alpha$ can be (wherever $tan(\theta - \alpha)$ & $tan(\theta + \alpha)$ both are defined)