Ребят нужна помощь срочно! Пожалуйста очень нужно, прошу!

Решите задачу:

$1)Sinx=\frac{\sqrt{2} }{2}\\\\x=(-1)^{n}arcSin\frac{\sqrt{2} }{2}+\pi n,nez\\\\x=(-1)^{n}\frac{\pi }{4}+\pi n,nez\\\\2)Cos2x=\frac{1}{2}\\\\2x=+-arcCos\frac{1}{2}+2\pi n,nez\\\\2x=+-\frac{\pi }{3} +2\pi n,nez\\\\x=+-\frac{\pi }{6}+\pi n,nez$

$3)tg2x=0\\\\2x=\pi n,nez\\\\x=\frac{\pi n }{2},nez\\\\4)4Sin^{2} x-3Sinx-1=0\\\\Sinx=m,-1\leq m\leq1\\\\4m^{2}-3m-1=0\\\\D=(-3)^{2}-4*4*(-1)=9+16=25=5^{2}\\\\m_{1}=\frac{3+5}{8}=1\\\\m_{2}=\frac{3-5}{8}=-\frac{1}{4}\\\\Sinx=1\\\\x=\frac{\pi }{2}+2\pi n,nez\\\\Sinx=-\frac{1}{4}\\\\x=(-1)^{n} arcSin(-\frac{1}{4} )+\pi n,nez\\\\x=(-1)^{n+1}arcSin\frac{1}{4}+\pi n,nez$

$5)Cos(\frac{5x}{3}+\frac{\pi }{6})=\frac{1}{2}\\\\ \frac{5x}{3}+\frac{\pi }{6}=+-arcCos\frac{1}{2}+2\pi n,nez\\\\\frac{5x}{3}+\frac{\pi }{6}=+-\frac{\pi }{3}+2\pi n,nez\\\\\frac{5x}{3} =+-\frac{\pi }{3}-\frac{\pi }{6}+2\pi n,nez\\\\x=+-\frac{\pi }{5}-\frac{\pi }{10}+\frac{6\pi n }{5},nez$

$6)Cos2\alpha=1-2Sin^{2}\alpha=1-2*(\frac{4}{5})^{2}=1-2*\frac{16}{25}=1-\frac{32}{25}=1-1\frac{7}{25}=-\frac{7}{25}=-0,28$