Question

Решение !!!!!!!!!!!!!!

Answer 1

$4sin^2x=sqrt2tgx; ,; ; ODZ:; xne frac{pi}{2}+pi n,; nin Z\\4sin^2x-frac{sqrt2sinx}{cosx}=0\\ frac{4sin^2xcdot cosx-sqrt2sinx}{cos}=0\\ frac{sqrt2sinxcdot (2sqrt2sinxcdot cosx-1)}{cosx} =0; ; to ; ; left { {{sqrt2sinxcdot (2sqrt2sinxcdot cosx)=0} atop {cosxne 0}} right. \\a); ; sinx=0; ,; ; x=pi k,; kin Z\\b); ; 2sqrt2sinxcdot cosx-1=0\\sqrt2cdot sin2x-1=0\\sin2x=frac{1}{sqrt2}; ,; ; ; ; frac{1}{sqrt2}=frac{sqrt2}{2}$

$2x=(-1)^{m}cdot frac{pi }{4}+pi m,; min Z\\x=(-1)^{m}cdot frac{pi}{8}+frac{pi m}{2},; min Z\\Otvet:; ; x=pi k,; kin Z; ; ;; ; x=(-1)^{m}frac{pi}{8}+frac{pi m}{2},; min Z ; .$