Question

Решить: 1) (sinx-3cosx)*(cosx+sinx)=1 2) (sinx-cosx)*(cosx+3sinx)=-1

Answer 1

$1) (sinx-3cosx)*(cosx+sinx)=1 \\ sinx*cosx+ sin^{2}x-3cos^{2}x- 3sinx*cosx-1=0\\ -2*sinx*cosx-3cos^{2}x-(1-sin^{2}x) =0\\ -2*sinx*cosx-3cos^{2}x-cos^{2}x=0\\ -2*sinx*cosx-4cos^{2}x=0\\ sinx*cosx+2cos^{2}x=0\\ cosx*(sinx+2cosx)=0\\ 1) cosx=0,\ then\ x= \frac{ \pi }{2}+ \pi *n,\ n?Z\ ;\ \\ 2)sinx+2cosx=0,/:cox \ ,\\ ctg+2=0 \ ,\\ tgx=-2,\ then\ x= arctg(-2)+ \pi *n,\ n?Z\ .\ \\$
$1) (sinx-cosx)*(cosx+3sinx)=-1 \\ sinx*cosx+ 3sin^{2}x-cos^{2}x- 3sinx*cosx+1=0\\ -2*sinx*cosx+3sin^{2}x+(1-cos^{2}x) =0\\ -2*sinx*cosx-3sin^{2}x+sin^{2}x=0\\ -2*sinx*cosx-2sin^{2}x=0\\ sinx*cosx+sin^{2}x=0\\ sinx*(cosx+sinx)=0\\ 1) sinx=0,\ then\ x=\pi *n,\ n?Z\ ;\ \\ 2)cosx+sinx=0,/:sinx \ ,\\ ctg+1=0 \ ,\\ ctgx=-1,\ then\ x= \frac{ \pi }{4}+ \pi *n,\ n?Z\ .\ \\$

Знак ? здесь заменяет знак принадлежности...