Đáp án: $S=0$
Giải thích các bước giải:
Ta có $\sin^4x-\cos^4x=(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)=\sin^2x-\cos^2x=-\cos2x$
$(7-\cos2x)(\sin^4x-\cos^4x)=0$
$\to (7-2\cos2x).\cos2x=0$
$\to \left[ \begin{array}{l}\cos2x=\dfrac{7}{2}(L) \\ \cos2x=0\end{array} \right.$
$\to x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
$-\pi<x<\pi\to k\in\left\{ -2;-1; 0; 1\right\}$
$\to x\in\left\{ \dfrac{-3\pi}{4}; \dfrac{-\pi}{4}; \dfrac{\pi}{4}; \dfrac{3\pi}{4}\right\}$
Tổng nghiệm: $S=0$