Question

1) Решите уравнение Cos 4x - √2 Cos 2x-1=0 (Нужно само решение!)
2) Найдите все корни, принадлежащие промежутку [-pi;pi]

Answer 1

$Cos4x-sqrt{2}Cos2x-1=0\\2Cos^{2}2x-1-sqrt{2}Cos2x-1=0\\2Cos^{2}2x-sqrt{2}Cos2x-2=0|:sqrt{2}\\sqrt{2}Cos^{2}2x-Cos2x-sqrt{2}=0\\Cos2x=m;-1leq mleq 1\\sqrt{2}m^{2} -m-sqrt{2} =0\\D=(-1)^{2}-4*sqrt{2}*(-sqrt{2})=1+8=9=3^{2}\\m_{1} =frac{1-3}{2sqrt{2} }=-frac{2}{2sqrt{2} }=-frac{sqrt{2} }{2}\\m_{2}=frac{1+3}{2sqrt{2} }=frac{4}{2sqrt{2}}=sqrt{2}>1\\Cos2x=-frac{sqrt{2} }{2}\\2x=pm arcCos(-frac{sqrt{2} }{2})+2pi n,nin z$

$2x=pmfrac{3pi }{4}+2pi n,nin z\\x=pmfrac{3pi }{8}+pi n,nin z\\1)x=frac{3pi }{8}+pi n,nin z\\n=-1\\x_{1} =-frac{5pi }{8}\\n=0\\x_{2}=frac{3pi }{8} \\2)x=-frac{3pi }{8}+pi n,nin z\\n=0\\x_{3}=-frac{3pi }{8}\\n=1\\x_{4}=frac{5pi }{8}$