Question

решите пожалуйста 
1+4cosx=cos2x sin4x+2cos^2x=1 3cosx+2tgx=0 2cos2x+4cosx=sin^2x

Answer 1

1)  $1+4cosx=cos2x; sin4x+2cos^2x=1; 3cosx+2tgx=0; 2cos2x+4cosx=sin^2x$
$1+4cosx=cos2x;$
$1+4cosx=2cos^2x-1;$
$2cos^2x-4cosx-2=0;cosx=t;t in [-1;1];$
$t^2-2t-1=0;D_1=2;t_1= 1+ sqrt{2}>1;t_2=1- sqrt{2};t_2 in [-1;1];$
2)  $sin4x+2cos^2x=1;$
$2sin2xcos2x+cos2x=0;$
$cos2x(2sin2x+1)=0;$
$cos2x=0;2x= frac{ pi }{2}+ pi n, n in Z;x= frac{ pi }{4}+ frac{ pi }{2} n, n in Z;$
$2sin2x+1=0;2sin2x=-1;sin2x=- frac{1}{2};$
$2x=(-1)^n(- frac{ pi }{6})+ pi n,n in Z; x=(-1)^{n+1}frac{ pi }{12}+ frac{ pi }{2} n,n in Z;$
3)  $3cosx+2tgx=0;$ $cosx neq 0;$
$3cosx+frac{2sinx}{cosx} =0; frac{3cos^2x+2sinx}{cosx}=0;$
$3cos^2x+2sinx=0;3(1-sin^2x)+2sinx=0;$
$3sin^2x-2sinx-3=0;sinx=t,t in [-1;1];$
$3t^2-2t-3=0;D_1=10;t_1= frac{1- sqrt{10}}{3};t_2=frac{1+ sqrt{10}}{3}>1;$
$sinx= frac{1- sqrt{10}}{3};x=(-1)^{n}arcsinfrac{1- sqrt{10}}{3}+ pi n,n in Z;$
4)  $2cos^2x+4cosx=sin^2x;$
$2cos^2x+4cosx=1-cos^2x;$
$3cos^2x+4cosx-1=0;cosx=t, t in [-1;1];$
$3t^2+4t-1=0;D_1=4+3=7;$
$t_1=frac{-2+ sqrt{7}}{3};t_2=frac{-2-sqrt{7}}{3}<-1$
$cosx= frac{-2+ sqrt{7}}{3};x=бarccos(frac{-2+ sqrt{7}}{3})+2 pi n, n in Z.$