Question

Cos3x-√3sin3x=2cosx люди добрые, решить. буду !

Answer 1

Объяснение:

$cos(3x)-\sqrt{3}sin(3x)=2cos(x)\\$

Умножаем обе части на 2

$2(\frac{cos(3x)}{2}+\frac{-\sqrt{3} cos(3x)}{2})=2cosx$

$2(cos( \frac{1}{2} cos(3x)-\frac{\sqrt{3}}{2} sin(3x))=2cosx$

$2(cos( \frac{\pi }{3}) cos(3x)-\frac{\sqrt{3}}{2} sin(3x))=2cosx$

Используем формулу:

$cos(x-y) = cos(x)cos(y) + sin(x)sin(y)$

$2(cos( \frac{\pi }{3}) + 3x)=2cosx$

$(cos( \frac{\pi }{3}) + 3x)=cosx$

$(cos( \frac{\pi }{3}) + 3x) - cosx = 0$

https://ru-static.z-dn.net/files/d2e/eab21d54911a5ef9afe4a113c36b6dc1.png

$sin(\frac{\pi+12 x}{6})(sin(\frac{\pi+6 x}{6})) = 0$

$sin(\frac{\pi+12 x}{6}) = 0\\sin(\frac{\pi+6 x}{6}) = 0$

$x = \frac{5\pi }{12} + \frac{\pi k}{2} \\x = \frac{5\pi }{6} + \pi k} \\k = Z$

Answer 2

$Cos3x-\sqrt{3}Sin3x=2Cosx\\\\2(\frac{1}{2}Cos3x-\frac{\sqrt{3} }{2}Sin3x)=2Cosx\\\\\frac{1}{2}Cos3x-\frac{\sqrt{3} }{2}Sin3x=Cosx\\\\Cos\frac{\pi }{3}Cos3x-Sin\frac{\pi }{3}Sin3x=Cosx\\\\Cos(\frac{\pi }{3}+3x)-Cosx=0$

$-2Sin\frac{\frac{\pi }{3}+3x+x }{2}Sin\frac{\frac{\pi }{3}+3x-x }{2}=0\\\\Sin(\frac{\pi }{6} +2x)Sin(\frac{\pi }{6}+x)=0\\\\1)Sin(\frac{\pi }{6} +2x)=0\\\\\frac{\pi }{6}+2x=\pi n,n\in z\\\\2x=-\frac{\pi }{6}+\pi n,n\n z\\\\x=-\frac{\pi }{12}+\frac{\pi n }{2},n\in z\\\\2)Sin(\frac{\pi}{6}+x)=0\\\\\frac{\pi }{6}+x=\pi n,n\in z\\\\x=-\frac{\pi }{6}+\pi n,n\in z\\\\Otvet:-\frac{\pi }{12}+\frac{\pi n }{2},n\in z;-\frac{\pi }{6}+\pi n,n\in z$