Question

Решите 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)  $3cosx+2tgx=0;$ $cosx \neq 0;$
$3cosx+\frac{2sinx}{cosx} =0; \frac{3cos^2x+2sinx}{cosx}=0;$
$3cos^2x+2sinx=0;3(1-sin^2x)+2sinx=0;$
$3sin^2x-2sinx-3=0;sinx=t,t \in [-1;1];$
$3t^2-2t-3=0;D_1=10;t_1= \frac{1- \sqrt{10}}{3};t_2=\frac{1+ \sqrt{10}}{3}1;$
$sinx= \frac{1- \sqrt{10}}{3};x=(-1)^{n}arcsin\frac{1- \sqrt{10}}{3}+ \pi n,n \in Z;$
4)  $2cos^2x+4cosx=sin^2x;$
$2cos^2x+4cosx=1-cos^2x;$
$3cos^2x+4cosx-1=0;cosx=t, t \in [-1;1];$
$3t^2+4t-1=0;D_1=4+3=7;$
$t_1=\frac{-2+ \sqrt{7}}{3};t_2=\frac{-2-\sqrt{7}}{3}<-1$
$cosx= \frac{-2+ \sqrt{7}}{3};x=бarccos(\frac{-2+ \sqrt{7}}{3})+2 \pi n, n \in Z.$