Question

решите уравнение 2+cos4x=3(cos^4x-sin^4x)

Answer 1

$displaystyle 2+cos4x=3(cos^4x-sin^4x)\\2+cos2(2x)=3(cos^2x-sin^2x)(cos^2x+sin^2x)\\2+(2cos^22x-1)=3(cos2x)*(1)\\2+2cos^22x-1-3cos2x=0\\2cos^22x-3cos2x+1=0\\cos2x=t\\2t^2-3t+1=0$
$displaystyle D=9-8=1\\t_{1.2}= frac{3pm 1}{4}\\t_1=1; t_2=1/2$

$displaystyle cos2x=1\\2x=2 pi n; nin Z\\x= pi n; nin Z$

$displaystyle cos2x=1/2\\2x=pm frac{ pi }{3}+2 pi n; nin Z\\x=pm frac{ pi }{6}+ pi n; nin Z$