Question

решите уравнение корень из cos^2x+cosx /sinx +1 =0

Answer 1

Ответ:

$\dfrac{cos^2x+cosx}{sinx+1}=0\ \ \ \Rightarrow \ \ \ \left\{\begin{array}{l}cos^2x+cosx=0\\sinx+1\ne 0\end{array}\right\ \ \left\{\begin{array}{l}cosx\, (cosx+1)=0\\sinx\ne -1\end{array}\right$

$\left\{\begin{array}{l}cosx=0\ \ ili\ \ cosx=-1\\sinx\ne -1\end{array}\right\ \ \left\{\begin{array}{l}\left[\begin{array}{l}x=\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\x=\pi +2\pi n\ ,\ n\in Z\end{array}\right\\x\ne -\dfrac{\pi}{2}+2\pi k\ ,\ k\in Z\end{array}\right\ \ \ \Rightarrow$

$Otvet:\ \ x=\dfrac{\pi}{2}+2\pi n\ \ ,\ \ x=\pi +2\pi k\ ,\ \ n,k\in Z\ .$