Question

2} sin(2x+\frac{\pi }{4} )-\sqrt{3} sinx=sin2x+1[/tex]

Answer 1

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

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