Question

Тригонометрическое уравнение

Answer 1

$\cos^2x-\dfrac{1}{2}\sin 2x+\cos x=\sin x\medskip\\\cos^2x-\sin x\cos x+\cos x-\sin x=0\medskip\\\cos x\left(\cos x-\sin x\right)+\left(\cos x-\sin x\right)=0\medskip\\\left(\cos x-\sin x\right)\left(\cos x+1\right)=0\medskip\\\left[\begin{gathered}\cos x-\sin x=0\\\cos x+1=0\end{gathered}\right.\Leftrightarrow\left[\begin{gathered}\mathrm{tg}~x=1\\\cos x=-1\end{gathered}\right.\Leftrightarrow\left[\begin{gathered}x=\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}\\x=\pi +2\pi m,~m\in\mathbb{Z}\end{gathered}$

Ответ. $x\in\left\{\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}\right\}\cup\left\{\pi+2\pi m,~m\in\mathbb{Z}\right\}$