One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

The number of solutions to $\sin \left(\pi \sin ^2 \theta\right)+\sin \left(\pi \cos ^2 \theta\right)=2 \cos \left(\frac{\pi}{2} \cos \theta\right)$ satisfying $0 \leq \theta \leq 2 \pi$ 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 $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0,\,\,2\pi ]$ is

The number of all possible values of $\theta$, where $0<\theta<\pi$, for which the system of equations

$ (y+z) \cos 3 \theta=(x y z) \sin 3 \theta $

$ x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z} $

$ (x y z) \sin 3 \theta=(y+2 z) \cos 3 \theta+y \sin 3 \theta$ have a solution $\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$ with $\mathrm{y}_0 \mathrm{z}_0 \neq 0$, is

The number of  $x \in  [0,2\pi ]$  for which $\left| {\sqrt {2\,{{\sin }^4}\,x\, + \,18\,{{\cos }^2}\,x}  - \,\sqrt {2\,{{\cos }^4}\,x\, + \,18\,{{\sin }^2}\,x} } \right| = 1$ is