Q. If $\sin 4 x \cos 4 x=1 / 2$ then $x=$______

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A

$\frac{n \pi}{2}+(-1)^{n} \frac{\pi}{16}$

B

$\frac{n \pi}{4}+\frac{\pi}{16}$

C

$\frac{nn}{12}+(-1)^{n} \frac{n}{12}$

D

$\frac{nn}{8}+(-1)^{n} \frac{n}{16}$

Solution:

given $\Rightarrow 2 \sin 4 x \cos 4 x=1$
$\Rightarrow \sin 8 x=1$
$\therefore 8 x=n \pi+(-1)^{n} \frac{\pi}{2}$
$\therefore x=\frac{n \pi}{8}+(-1)^{n} \frac{\pi}{16}$

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