If $\cos \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1$, then the general value of $\theta $ is
$2n\pi + \frac{\pi }{4}$
$(2n + 1)\,\pi + \frac{\pi }{4}$
$n\pi + \frac{\pi }{4}$
$n\pi \pm \frac{\pi }{4}$
The general solution of the equation $sin^{100}x\,-\,cos^{100} x= 1$ is
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)
$(A)$ $\frac{\pi}{4}$ $(B)$ $\frac{\pi}{6}$ $(C)$ $\frac{\pi}{12}$ $(D)$ $\frac{5 \pi}{12}$
Find the solution of $\sin x=-\frac{\sqrt{3}}{2}$
If $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ then $x = $
The only value of $x$ for which ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$ holds, is