Question

Решить уравнение: $(sin2x+\sqrt{3}cos2x)^2=5+cos(\frac{\pi}{6}-2x)$
С подробным, пошаговым решением.

Answer 1

Ответ:

$(sin2x+\sqrt3\, cos2x)^2=5+cos\Big(\dfrac{\pi}{6}-2x\Big)\\\\\star \ \ (sin2x+\sqrt3\, cos2x)=2\cdot \Big(\dfrac{1}{2}\cdot sin2x+\dfrac{\sqrt3}{2}\cdot cos2x\Big)=\\\\=2\cdot \Big(sin\dfrac{\pi}{6}\cdot sin2x+cos\dfrac{\pi}{6}\cdot cos2x\Big)=2\cdot cos\Big(\dfrac{\pi}{6}-2x\Big)\ \ \star \\\\\\4\cdot cos^2\Big(\dfrac{\pi}{6}-2x\Big)-cos\Big(\dfrac{\pi}{6}-2x\Big)-5=0\ \ ,\ \ \ \ t=cos\Big(\dfrac{\pi}{6}-2x\Big)\ ,$

$t\in [-1\, ;\, 1\, ]\ \ ,\ \ 4t^2-t-5=0\ \ ,\ \ \ D=1+80=81\ \ ,\\\\t_1=-1\ ,\ \ t_2=\dfrac{5}{4}=1,25>1\\\\cos\Big(\dfrac{\pi}{6}-2x\Big)=-1\\\\\\\dfrac{\pi}{6}-2x=\pi +2\pi n\ \ ,\ \ \ 2x=\dfrac{\pi}{6}-\pi -2\pi n\ \ ,\ \ \ 2x=-\dfrac{5\pi}{6}-2\pi n\ ,\ n\in Z\ ,\\\\\\x=-\dfrac{5\pi}{12}-\pi n\ ,\ n\in Z\\\\\\Otvet:\ \ x=-\dfrac{5\pi}{12}-\pi n\ ,\ n\in Z\ .$