Question

решить уравнение
cos2x+sinx= -1

3cos^2(x)+3sinxcosx=0
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Answer 1

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

$2)\ \ 3cos^2x+3\, sinx\cdot cosx=0\\\\3\, cosx\cdot (cosx+sinx)=0\\\\a)\ \ cosx=0\ ,\ \ x=\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\\\b)\ \ cosx+sinx=0\ \Big|:cosx\ne 0\\\\1+tgx=0\ \ ,\ \ \ tgx=-1\ \ ,\ \ \ x=-\dfrac{\pi}{4}+\pi k\ ,\ k\in Z\\\\\\Otvet:\ \ x_1=\dfrac{\pi}{2}+\pi n\ \ ,\ \ x_2=-\dfrac{\pi}{4}+\pi k\ \ ,\ k,n\in Z\ .$