All the pairs $(x, y)$ that satisfy the inequality ${2^{\sqrt {{{\sin }^2}{\kern 1pt} x - 2\sin {\kern 1pt} x + 5} }}.\frac{1}{{{4^{{{\sin }^2}\,y}}}} \leq 1$ also Satisfy the equation
$2\left| {\sin \,x} \right| = 3\sin \,y$
$\sin \,x = \left| {\sin \,y} \right|$
$2\,sin\, x = sin\, y$
$sin\, x = 2\, sin\, y$
The set of values of $x$ satisfying the equation,${2^{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 4}}} \right)}}$ $- 2$${\left( {0.25} \right)^{\frac{{{{\sin }^2}\,\left( {x\,\, - \,\,{\textstyle{\pi \over 4}}} \right)}}{{\cos \,\,2x}}}}$ $+ 1 = 0$, is :
One of the solutions of the equation $8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta=0$ lies in the interval
The value of $\theta $ in between ${0^o}$ and ${360^o}$ and satisfying the equation $\tan \theta + \frac{1}{{\sqrt 3 }} = 0$ is equal to
If $sin^2x + sinx \,cosx -6cos^2x = 0$ and $-\frac{\pi}{2} < x < 0$, then the value of $cos2x$, is
All possible values of $\theta \in[0,2 \pi]$ for which $\sin 2 \theta+\tan 2 \theta>0$ lie in