Помогите пожалста!! решите уравнение: sin^2(2x+pi/4)=1/4

[latex]\sin^2{\left(2x+\frac{\pi}{4} \right)}=\frac{1}{4},\\\\\frac{1-\cos{\left(2\left(2x+\frac{\pi}{4}\right)\right)}}{2}=\frac{1}{4},\\\\\frac{1-\cos{\left(4x+\frac{\pi}{2}\right)}}{2}=\frac{1}{4}\ |\cdot2,\\\\1-\cos{\left(4x+\frac{\pi}{2}\right)}}=\frac{1}{2},\\\\-\cos{\left(4x+\frac{\pi}{2}\right)}}=\frac{1}{2}-1,\\\\-\cos{\left(4x+\frac{\pi}{2}\right)}}=-\frac{1}{2},\\\\\sin{4x}=-\frac{1}{2},\\\\-1\le-\frac{1}{2}\le1\ \Longrightarrow\ 4x=\left(-1\right)^n\arcsin{\left(-\frac{1}{2}\right)}+\pi n,\ n\in Z,[/latex] [latex]4x=\left(-1\right)^n\cdot\frac{7\pi}{6}+\pi n,\ n\in Z,\\\\x=\left(-1\right)^n\cdot\frac{7\pi}{6\cdot4}+\frac{\pi n}{4},\ n\in Z,\\\\x=\left(-1\right)^n\cdot\frac{7\pi}{24}+\frac{\pi n}{4},\ n\in Z.\\\\ OTBET:\ x=\left(-1\right)^n\cdot\frac{7\pi}{24}+\frac{\pi n}{4},\ n\in Z.[/latex]