Question

Решить уравнения 1) sin^2x+cosx+1=0 2)3sin^2x+2sinxcosx-5cos^2x=0 3)sinx-cosx=0 4)sin3x*ctgx-3sin3x=0

Answer 1

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

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

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

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