Question

Пожалуйста помогите решить

Answer 1

$4)Cos(\frac{\pi }{4}+2x)=0\\\\\frac{\pi }{4}+2x=\frac{\pi }{2}+\pi n,n\in Z\\\\2x=\frac{\pi }{2}-\frac{\pi }{4}+\pi n,n\in Z\\\\2x=\frac{\pi }{4}+\pi n,n\in Z\\\\x=\frac{\pi }{8}+\frac{\pi n}{2},n\in Z\\\\Otvet:\boxed{\frac{\pi }{8}+\frac{\pi n }{2},n\in Z}$

$5)\sqrt[3]{5-2\sqrt{6}}*\sqrt[3]{5+2\sqrt{6}}=\sqrt[3]{(5-2\sqrt{6})(5+2\sqrt{6})}=\sqrt[3]{5^{2}-(2\sqrt{6})^{2}}=\sqrt[3]{25-24}=\sqrt[3]{1}=1\\\\Otvet:\boxed{1}$