Đáp án:

$C. 62\pi$

Giải thích các bước giải:

$2\sin^2\left(x + \dfrac{\pi}{4}\right) + 1 = \cos^2x$

$\Leftrightarrow 2.\dfrac{1 - \cos\left(2x + \dfrac{\pi}{2}\right)}{2} + 1 - \cos^2x = 0$

$\Leftrightarrow 1 + \sin2x - \cos2x = 0$

$\Leftrightarrow \sqrt2\sin\left(2x - \dfrac{\pi}{4}\right) = -1$

$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \dfrac{\pi}{2}+ k\pi\end{array}\right. \,\,\,\,(k\in\Bbb Z)$

Do $x \in (0;8\pi),\, k \in \Bbb Z$, ta được:

$\star \, 0 < \dfrac{\pi}{4} + k\pi < 8\pi$

$\Leftrightarrow - \dfrac{1}{4} < k < \dfrac{31}{4}$

$\Rightarrow k = \left\{0;1;2;3;4;5;6;7\right\}$

$\star \, 0 < \dfrac{\pi}{2} + k\pi < 8\pi$

$\Leftrightarrow -\dfrac{1}{2} < k < \dfrac{15}{2}$

$\Rightarrow k = \left\{0;1;2;3;4;5;6;7\right\}$

$\Rightarrow S = \left\{\dfrac{\pi}{4}.8 + [(7+1).3 +4]\pi\right\} + \left\{\dfrac{\pi}{2}.8 + [(7 +1).3 +4]\pi \right\} = 2\pi + 28\pi + 4\pi + 28\pi = 62\pi$