Question

решить уравнение
cos2x+sinx= -1
​

Answer 1

Ответ:

$cos2x+sinx=-1\\\\\star \ \ \underline{cos2x}=cos^2x-sin^2x=(1-sin^2x)-sin^2x=\underline {1-2sin^2x}\ \ \star \\\\1-2sin^2x+sinx=-1\\\\2sin^2x-sinx-2=0\\\\t=sinx\ ,\ \ t\in [-1\, ;\, 1\, ]\ \ ,\ \ \ 2t^2-t-2=0\ \ ,\ \ D=17\ ,\ t_{1,2}=\dfrac{1\pm \sqrt{17}}{4}\\\\a)\ \ sinx=\dfrac{1+\sqrt{17}}{4}\approx 1,28>1\ \ \to \ \ \ x\in \varnothing \ \ ,\\\\\\b)\ \ sinx=\dfrac{1-\sqrt{17}}{4}\approx -0,78\ \ ,\ \ x=(-1)^{n}\cdot arcsin\dfrac{1-\sqrt{17}}{4}+\pi k\ ,\ k\in Z$

$Otvet:\ \ x=(-1)^{n}\cdot arcsin\dfrac{1-\sqrt{17}}{4}+\pi k\ ,\ k\in Z\ .$