Question

ПОМОГИТЕ ! СРОЧНО
!!!!!!$sqrt{2} sin(2x+frac{pi }{4} )-sqrt{3} sinx=sin2x+1$

Answer 1

$sqrt2, sin(2x+frac{pi}{4})-sqrt3, sinx=sin2x+1\\sqrt2, (sin2xcdot cosfrac{pi}{4}+cos2xcdot sinfrac{pi}{4})-sqrt3, sinx-sin2x-1=0\\ sqrt2cdot frac{sqrt2}{2}cdot (sin2x+cos2x)-sqrt3, sinx-sin2x-1=0\\underline {sin2x}+cos2x-sqrt3, sinx-underline {sin2x}-1=0\\cos^2x-sin^2x-sqrt3, sinx-(sin^2x+cos^2x)=0\\-2sin^2x-sqrt3, sinx=0\\sinxcdot (2sinx+sqrt3)=0\\a); sinx=0; ; ,; ; x=pi n,; nin Z\\b); ; 2sinx+sqrt3=0; ; ,; ; sinx=-frac{sqrt3}{2}; ,$

$x=(-1)^{n}cdot (-frac{pi}{3})+pi n=(-1)^{n+1}cdot frac{pi}{3}+pi n,; nin Z\\Otvet:; ; x=pi n; ,; ; x=(-1)^{n+1}cdot frac{pi }{3}+pi n,; nin Z; .$