Question

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Answer 1

$4Sin^{4} 2x+3Cos4x=1\\\\4Sin^{4} 2x+3\cdot(1-2Sin^{2} 2x)=1\\\\4Sin^{4} 2x+3-6Sin^{2} 2x-1=0\\\\4Sin^{4} 2x-6Sin^{2} 2x+2=0 \ |:2\\\\2Sin^{4} 2x-3Sin^{2} 2x+1=0\\\\Sin^{2}2x=m \ \ \Rightarrow \ \ Sin^{4} 2x=m^{2} \\\\2m^{2} -3m+1=0\\\\D=(-3)^{2} -4\cdot 2\cdot 1=9-8=1\\\\\\m_{1} =\dfrac{3-1}{4} =\dfrac{1}{2} \\\\\\m_{2} =\dfrac{3+1}{4}=1$

$1)\\\\Sin^{2}2x=\dfrac{1}{2} \\\\\\Sin2x_{1} =\dfrac{1}{\sqrt{2} } \\\\\\2x_{1} =(-1)^{n} arcSin\dfrac{1}{\sqrt{2} } +\pi n,n\in Z\\\\\\2x_{1}= (-1)^{n} \dfrac{\pi }{4} +\pi n,n\in Z\\\\\\\boxed{x_{1} =(-1)^{n} \dfrac{\pi }{8} +\dfrac{\pi n}{2} ,n\in Z}$

$Sin2x_{2} =-\dfrac{1}{\sqrt{2} } \\\\\\2x_{2} =(-1)^{n} arcSin\Big(-\dfrac{1}{\sqrt{2} }\Big) +\pi n,n\in Z\\\\\\2x_{2}= (-1)^{n+1} \dfrac{\pi }{4} +\pi n,n\in Z\\\\\\\boxed{x_{2} =(-1)^{n+1} \dfrac{\pi }{8} +\dfrac{\pi n}{2} ,n\in Z}\\\\\\2)Sin^{2} 2x=1\\\\\\Sin2x_{3} =1\\\\\\2x_{3} =\dfrac{\pi }{2} +2\pi n,n\in Z\\\\\\\boxed{x_{3} =\dfrac{\pi }{4} +\pi n,n\in Z}\\\\\\Sin2x_{4} =-1\\\\\\2x_{4} =-\dfrac{\pi }{2} +2\pi n,n\in Z\\\\\\\boxed{x_{4}=-\dfrac{\pi }{4} +\pi n,n\in Z }$