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)

For $0<\theta<\frac{\pi}{2}$, the solution(s) of $\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}$ is(are)

$(A)$ $\frac{\pi}{4}$ $(B)$ $\frac{\pi}{6}$ $(C)$ $\frac{\pi}{12}$ $(D)$ $\frac{5 \pi}{12}$

The set of angles btween $0$ & $2\pi $ satisfying the equation $4\, cos^2 \, \theta - 2 \sqrt 2 \, cos \,\theta - 1 = 0$ is

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

The equation $\sqrt 3 \sin x + \cos x = 4$ has