Question

Помогите пожалуйста: cos4x-sin2x+1=0

Answer 1

$Cos4x-Sin2x+1=0\\\\1-2Sin^{2}2x-Sin2x+1=0\\\\2Sin^{2}2x+Sin2x-2=0\\\\Sin2x=m,-1\leq m \leq1\\\\2m^{2}+m-2=0\\\\D=1^{2}-4*2*(-2)=1+16=17\\\\m_{1}=\frac{-1+\sqrt{17} }{4} \\\\m_{2}=\frac{-1-\sqrt{17} }{4}<-1-neyd\\\\Sin2x=\frac{\sqrt{17}-1 }{4}\\\\2x=(-1)^{n}arc Sin\frac{\sqrt{17}-1 }{4}+\pi n,n\in Z\\\\x=\frac{1}{2}*(-1)^{n}arc Sin\frac{\sqrt{17}}{4}+\frac{\pi n }{2},n\in Z$