$\sin 4x={{\sin }^{4}}x-{{\cos }^{4}}x\to \sin 4x=\left( \underbrace{{{\sin }^{2}}x {{\cos }^{2}}x}_{1} \right)\left( \underbrace{{{\sin }^{2}}x-{{\cos }^{2}}x}_{-\cos 2x} \right)$
$\Rightarrow 2\sin 2x\cos x=-\cos 2x\Rightarrow 2\sin 2x\cos 2x \cos 2x=0\Rightarrow \cos 2x\left( 2\sin x 1 \right)=0$
مجموع جوابها: $\frac{\pi }{4} \frac{3\pi }{4} \frac{11\pi }{12} \frac{7\pi }{12}=\frac{30\pi }{12}=\frac{5\pi }{2}$