sinx-2cosx=2 решите даю 40 баллов. Created by zumrat62. algebra-ru

sinx-2cosx=2 решите даю 40 баллов...

армения20171

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

$x=(-1)^{n}\cdot arcsin\frac{2}{\sqrt5}+arctg2+\pi n\; ,\; n\in Z$

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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)

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