Question

$2cos(2x)= \sqrt{6} (cos(x)-sin(x))$

Answer 1

$2cos(2x)= \sqrt{6} (cos(x)-sin(x))\\2(cos^2(x)-sin^2(x))= \sqrt{6} (cos(x)-sin(x))\\2(cos(x)-sin(x))(cos(x)+sin(x))= \sqrt{6} (cos(x)-sin(x))\\(cos(x)-sin(x))(2(cos(x)+sin(x))- \sqrt{6})=0\\ \sqrt{2} cos(x+ \frac{\pi}{4} ) (2 \sqrt{2}cos(x- \frac{\pi}{4} )- \sqrt{6})=0\\cos(x+ \frac{\pi}{4} )(2cos(x- \frac{\pi}{4} )- \sqrt{3} )=0\\cos(x+ \frac{\pi}{4} )=0\\x+ \frac{\pi}{4} = \frac{\pi}{2} +k \pi\\x= \frac{\pi}{2} - \frac{\pi}{4} +k \pi\\x= \frac{\pi}{4} +k \pi$
$2cos(x- \frac{\pi}{4} )- \sqrt{3} =0\\cos(x- \frac{\pi}{4} )= \frac{ \sqrt{3} }{2} \\x- \frac{\pi}{4} =+- \frac{\pi}{6} +2\pi k\\x= \frac{\pi}{4} +- \frac{\pi}{6} +2\pi k\\ \frac{\pi}{4}-\frac{\pi}{6}=\frac{2\pi}{24} =\frac{\pi}{12} \\\\\frac{\pi}{4}+\frac{\pi}{6}=\frac{10\pi}{24} =\frac{5\pi}{12}$
Не забываем k∈Z
Ответ: $x=\frac{\pi}{4}+\pi k\\x=\frac{\pi}{12} +2\pik\\\\x=\frac{5\pi}{12} +2\pi k$

Answer 2

$2cos2x= \sqrt{6} (cosx-sinx) \\ \\ 2(cos^2x-sin^2x) - \sqrt{6} (cosx-sinx)=0 \\ \\ 2(cosx-sinx)(cosx+sinx) - \sqrt{6} (cosx-sinx)=0 \\ \\ (cosx-sinx)(2(cosx+sinx) - \sqrt{6})=0$

Приравниваем каждый множитель к нулю:
cosx - sinx = 0 (однородное уравнение 1-ой степени; делим на cosx ≠ 0)
1 - tgx = 0
tgx = 1
$x = \frac{\pi}{4} +\pi k$
k ∈ Z
_____________________
$2(cosx+sinx) - \sqrt{6}=0 \\ cosx+sinx= \frac{ \sqrt{6} }{2} \\ cosx+sinx= \sqrt{ \frac{3}{2} }$
Возведём в квадрат обе части:
$(cosx+sinx)^2= \frac{3}{2} \\ cos^2x+2sinxcosx+sin^2x= \frac{3}{2} \\ 1+sin2x= \frac{3}{2}\\ sin2x= \frac{1}{2} \\ \\ 2x= \frac{\pi}{6} +2\pi k \\ 2x= \frac{5\pi}{6} +2\pi k \\ \\ x= \frac{\pi}{12} +\pi k \\ x= \frac{5\pi}{12} +\pi k$
k ∈ Z

$OTBET: \frac{\pi}{4} +\pi k; \frac{\pi}{12} +\pi k; \frac{5\pi}{12} +\pi k; k \in Z$