The sum of the solutions of the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4}$ in the interval $(0, 2\pi)$ is

  • A
    $4\pi$
  • B
    $\pi$
  • C
    $2\pi$
  • D
    $3\pi$

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The value of $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$ is

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Match the items of List-$I$ to the items of List-$II$:
List-$I$List-$II$
$A$. The period of $\sin^2 x$ is$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=|\sin x|$ and $y=1$ in $(0, \pi)$$IV$. $\frac{3\pi}{2}$
$V$. $\pi$

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$,for $x \in R$,then $f(2002)$ is equal to

Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x = \frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x)$ is equal to

$4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta = $

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