Question

решить уравнения
1) sin^2x+cosx+1=0
2)3sin^2x+2sinxcosx-5cos^2x=0
3)$sqrt{3}$sinx-cosx=0
4)sin3x*ctgx-3sin3x=0

Answer 1

$1) sin^2x+cosx+1=0 \ \ (1-cos^2x)+cosx+1=0 \ \ -cos^2x+cosx+2=0 \ \ cos^2x-cosx-2=0 \ \ cosx=y \ \ -1leq y leq 1 \ \ y^2-y-2=0 \ \ D=1+4*2=9=3^2 \ \ y_1= frac{1-3}{2}=-1 \ \ y_2= frac{1+3}{2}=2 textgreater 1 \ \ cosx=-1 \ \ x= pi +2 pi n, nin Z$

$2)3sin^2x+2sinxcosx-5cos^2x=0 \ \ 3sin^2x-3sinxcosx+5sinxcosx-5cos^2x=0 \ \ 3sinx(sinx-cosx)+5cosx(sinx-cosx)=0 \ \ (3sinx+5cosx)(sinx-cosx)=0 \ \ 3sinx+5cosx=0 :cosx \ \ 3 frac{sinx}{cosx}+5=0 \ \ 3tgx=-5 \ \ tgx=- frac{5}{3} \ \ x=arctg(- frac{5}{3})+ pi *n, nin Z \ \ sinx-cosx=0 :cosx \ \ frac{sinx}{cosx}-1=0 \ \ tgx=1 \ \ x= frac{ pi }{4} + pi n, nin Z$

$3) sqrt{3} sinx-cosx=0 : cosx \ \ sqrt{3} frac{sinx}{cosx}-1=0 \ \ sqrt{3}tgx=1 \ \ tgx= frac{ sqrt{3} }{3} \ \ x= frac{ pi }{6}+ pi n, nin Z$

$4)sin3x*ctgx-3sin3x=0 \ \ sin3x(ctgx-3)=0 \ \ sin3x=0 \ \ 3x= pi n, nin Z \ \ x= frac{ pi n }{3} \ \ ctgx=3 \ \ x=arcctg3+ pi n, nin Z$