${{\operatorname{Sin}}^{4}}x-{{\operatorname{Cos}}^{4}}x=(\underbrace{{{\operatorname{Sin}}^{2}}x-{{\operatorname{Cos}}^{2}}x}_{-\operatorname{Cos}2x})(\underbrace{{{\operatorname{Sin}}^{2}}x+{{\operatorname{Cos}}^{2}}x}_{1})=-\operatorname{Cos}2x$  $\operatorname{Sin}4x=\operatorname{Cos}2x=\operatorname{Sin}(\frac{3\pi }{2}+2x)\Rightarrow \left\{ _{4x=2k\pi +\pi -(\frac{3\pi }{2}+2x)}^{4x=2k\pi +(\frac{3\pi }{2}+2x)}\Rightarrow \left\{ _{x=\frac{k\pi }{3}-\frac{\pi }{12}=\frac{4\pi -\pi }{12}}^{x=k\pi +\frac{3\pi }{4}} \right. \right.,k\in Z$  جواب‌های واقع در $\left[ 0,\pi  \right]$ $\left\{ _{x=\frac{\pi }{4},\frac{7\pi }{12},\frac{11\pi }{12}}^{x=\frac{3\pi }{4}} \right.$  بنابراین مجموع جواب‌ها برابر $\frac{3\pi }{4}+\frac{\pi }{4}+\frac{7\pi }{12}+\frac{11\pi }{12}=\frac{5\pi }{2}$ است.