If sum of all the solutions of the equation $8\cos x \cdot \left( {\cos \left( {\frac{\pi }{6} + x} \right) \cdot \cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) = 1$ in $\left[ {0,\pi } \right]$ is $k\pi $then $k$ is equal to :
$\frac{{13}}{9}$
$\frac{8}{9}$
$\frac{{20}}{9}$
$\frac{2}{3}$
Solve $\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)$
The general solution of $\tan 3x = 1$ is
If $2{\sin ^2}\theta = 3\cos \theta ,$ where $0 \le \theta \le 2\pi $, then $\theta = $
If $\sec 4\theta - \sec 2\theta = 2$, then the general value of $\theta $ is
Let $f:[0,2] \rightarrow R$ be the function defined by
$f ( x )=(3-\sin (2 \pi x )) \sin \left(\pi x -\frac{\pi}{4}\right)-\sin \left(3 \pi x +\frac{\pi}{4}\right)$
If $\alpha, \beta \in[0,2]$ are such that $\{x \in[0,2]: f(x) \geq 0\}=[\alpha, \beta]$, then the value of $\beta-\alpha$ is. . . . . . . . .