The solution of the equation $4{\cos ^2}x + 6$${\sin ^2}x = 5$
$x = n\pi \pm \frac{\pi }{2}$
$x = n\pi \pm \frac{\pi }{4}$
$x = n\pi \pm \frac{{3\pi }}{2}$
None of these
Let $X=\{x \in R: \cos (\sin x)=\sin (\cos x)\} .$ The number of elements in $X$ is
$sin^{2n}x + cos^{2n}x$ lies between
If $(2\cos x - 1)(3 + 2\cos x) = 0,\,0 \le x \le 2\pi $, then $x = $
The numbers of solution $(s)$ of the equation $\left( {1 - \frac{1}{{2\,\sin x}}} \right){\cos ^2}\,2x\, = \,2\,\sin x\, - \,3\, + \,\frac{1}{{\sin x}}$ in $[0,4\pi ]$ is
The solution of the equation $\left| {\,\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }&{\cos \theta }\\{ - \sin \theta }&{\cos \theta }&{\sin \theta }\\{ - \cos \theta }&{ - \sin \theta }&{\cos \theta }\end{array}\,} \right| = 0$, is