Question

$begin{cases} 2cos^2x+1=2sqrt{2}sinx \ sinxcosy=frac{1}{2} end{cases}$

Answer 1

$begin{cases} 2cos^2x+1=2sqrt{2}sinx \ sinxcosy=frac{1}{2} end{cases} \ sinx=a \ siny=b \ cos^2x=1-a^2 begin{cases} 2(1-a^2)+1=2sqrt{2}a \ ab=frac{1}{2} end{cases} \ begin{cases} 2-2a^2+1=2sqrt{2}a \ ab=frac{1}{2} end{cases} 2a^2+2sqrt{2}a-3=0 \ D=2+6=8 \ a neq frac{-sqrt{2}-2sqrt{2}}{2} = frac{-3sqrt{2}}{2} <-1 \ a =frac{-sqrt{2}+2sqrt{2}}{2} = frac{sqrt{2}}{2} \ sinx= frac{sqrt{2}}{2}$
$x=(-1)^k frac{pi}{4} +pi k, kin Z \ frac{ sqrt{2} }{2} cosy= frac{1}{2} \ cosy=frac{ sqrt{2} }{2} \ y=pm frac{pi}{4} +2pi n , nin Z$
Ответ: $x=(-1)^k frac{pi}{4} +pi k, kin Z \ y=pm frac{pi}{4} +2pi n , nin Z$