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

Answer 2

2) 2cos(x -π/4) - √3 =0 ;

Answer 3

x =π/12 +2πk , x =5π/12 +2πk

Answer 4

так у него правильно?

Answer 5

да , только в ответе x = π/12 +2πk (вместо x = π/12 +2 ) → описка

Answer 6

ясно.А почему он не исправит?