Question

sinx-2cosx=2 решите даю 40 баллов

Answer 1

sinx-2cosx=2

√5*(1/√5*sinx-2/√5*cosx)=2
sinb=1/√5;cosb=2/√5
cos(x-b)=-2/√5
x-b=±(arccos(-2/√5))+2πk

x=arccos2/√5±(π-arccos2/√5)+2πk;k€Z
x1=π+2πk=π(1+2k)
x2=2arccos2/√5-π+2πk=
2arccos2/√5+π(2k-1)

Answer 2

$sinx-2cosx=2; |:sqrt5\\frac{1}{sqrt5}, sinx-frac{2}{sqrt5}, cosx=frac{2}{sqrt5}\\Tak; kak; ; (frac{1}{sqrt5})^2+(frac{2}{sqrt5})^2=frac{5}{5}=1; ,; to; ; frac{1}{sqrt5}=cosa; ,; frac{2}{sqrt5}=sina; Rightarrow \\tga=frac{sina}{cosa}=2; ; to ; ; a=arctg2\\cosacdot sinx-sinacdot cosx=frac{2}{sqrt5}\\sin(x-a)=frac{2}{sqrt5}\\x-a=(-1)^{n}cdot arcsinfrac{2}{sqrt5}+pi n; ,; nin Z\\x=(-1)^{n}cdot arcsinfrac{2}{sqrt5}+a+pi n; ,; nin Z$

$x=(-1)^{n}cdot arcsinfrac{2}{sqrt5}+arctg2+pi n; ,; nin Z$