The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is
$8$
$5$
$6$
$7$
If ${\sin ^2}\theta = \frac{1}{4},$ then the most general value of $\theta $ is
The sum of solutions in $x \in (0,2\pi )$ of the equation, $4\cos (x).\cos \left( {\frac{\pi }{3} - x} \right).\cos \left( {\frac{\pi }{3} + x} \right) = 1$ is equal to
$sin 3\theta = 4 sin\, \theta \,sin \,2\theta \,sin \,4\theta$ in $0\, \le \,\theta\, \le \, \pi$ has :
The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for
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 :