Question

cos3x-sinx=корень из 3 (cosx-sin3x)

Answer 1

$cos3x-sin x= sqrt{3} (cos x-sin 3x)\ cos3x-sin x=sqrt{3} cos x-sqrt{3} sin3x\ cos3x+sqrt{3} sin3x=sin x+sqrt{3} cos x\ sqrt{(sqrt{3} )^2+1^2} sin(3x+arcsin frac{1}{ sqrt{(sqrt{3} )^2+1^2} } )= sqrt{(sqrt{3} )^2+1^2} sin(x+ arcsinfrac{sqrt{3}}{sqrt{(sqrt{3})^2+1^2}}$

$2sin(3x+ frac{pi}{6} )=2sin(x+frac{pi}{3}) \ \ sin(3x+frac{pi}{6} )=sin(x+frac{pi}{3}) \ \ sin(3x+frac{pi}{6} )-sin(x+frac{pi}{3})=0\ \ 2cos frac{3x+frac{pi}{6} +x+frac{pi}{3} }{2} sin frac{3x+frac{pi}{6} -x-frac{pi}{3} }{2} =0\ \ 2cos frac{4x+frac{pi}{2} }{2} sin frac{2x-frac{pi}{6} }{2} =0\ \ 2cos(2x+frac{pi}{4} )sin(x-frac{pi}{12} )=0$

Произведение равно нулю, если один из множителей равен нулю

$cos(2x+frac{pi}{4} )=0\ \ 2x+frac{pi}{4}=frac{pi}{2}+ pi n,n in Z\ \ 2x=frac{pi}{2}-frac{pi}{4}+ pi n,n in Z \ \ 2x=frac{pi}{4}+ pi n,n in Z\ \ boxed{x_1= frac{pi}{8} +frac{pi n}{2},n in Z}$

$sin(x-frac{pi}{12})=0\ \ x-frac{pi}{12}= pi k,k in Z\ \ boxed{x_2=frac{pi}{12}+ pi k,k in Z}$