Question

Решите уравнение 1+cos2x+sin2x=0 и найдите корни из промежутка (-3П; -2П)

Answer 1

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Answer 2

а)

$1+\cos 2x+\sin 2x=0\medskip\\\sin^2x+\cos^2x+\cos^2x-\sin^2x+2\sin x\cos x=0\medskip\\2\cos^2x+2\sin x\cos x=0\medskip\\\cos x\left(\cos x+\sin x\right)=0\medskip\\1)~\cos x=0\medskip\\x=\dfrac{\pi}{2}+\pi k,~k\in\mathbb{Z}\medskip\\2)~\cos x+\sin x=0 \mid\div\cos x\neq 0\medskip\\1+\mathrm{tg}~x=0\medskip\\\mathrm{tg}~x=-1\medskip\\x=-\dfrac{\pi}{4}+\pi t,~t\in\mathbb{Z}$

б)

$\left(-3\pi;-2\pi\right)\medskip\\1)~-3\pi<\dfrac{\pi}{2}+\pi m<-2\pi;~m\in\mathbb{Z}\medskip\\-3{,}5<m<-2{,}5;~m\in\mathbb{Z}\Rightarrow m=-3\medskip\\x_1=\dfrac{\pi}{2}-3\pi=-\dfrac{5\pi}{2}\medskip\\2)~-3\pi<-\dfrac{\pi}{4}+\pi k<-2\pi;~k\in\mathbb{Z}\medskip\\-2{,}75<k<-1{,}75;~k\in\mathbb{Z}\Rightarrow k=-2\medskip\\x_2=-\dfrac{\pi}{4}-2\pi=-\dfrac{9\pi}{4}$

Ответ. а) $x=\dfrac{\pi}{2}+\pi m,~m\in\mathbb{Z} \vee x=-\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}$; б) $\left\{-\dfrac{5\pi}{2};-\dfrac{9\pi}{4}\right\}$