Решите уравнение sin2x−2sinx+2cosx−2=0\sin 2x - 2\sin x + 2\cos x - 2 = 0sin2x−2sinx+2cosx−2=0. Источник: Досрочная волна ЕГЭ-2015 Показать решение Решение: 2sinxcosx−2sinx+2cosx−2=02\sin x \cos x - 2\sin x + 2\cos x - 2 = 02sinxcosx−2sinx+2cosx−2=0 2sinx(cosx−1)+2(cosx−1)=02\sin x (\cos x - 1) + 2(\cos x - 1) = 02sinx(cosx−1)+2(cosx−1)=0 2(cosx−1)(sinx+1)=0⇔(cosx−1)(sinx+1)⇔2(\cos x - 1)(\sin x + 1) = 0 \Leftrightarrow (\cos x - 1)(\sin x + 1) \Leftrightarrow2(cosx−1)(sinx+1)=0⇔(cosx−1)(sinx+1)⇔ ⇔[sinx=−1cosx=1⇔[x=−π2+2πkx=2πk,k∈Z\Leftrightarrow \left[ \begin{aligned} &\sin x = -1 \\ &\cos x = 1 \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} & x = -\frac{\pi}{2} + 2\pi k\\ & x = 2\pi k \end{aligned} \right., k\in Z⇔[sinx=−1cosx=1⇔[x=−2π+2πkx=2πk,k∈Z Ответ: x=2πk; x=−π2+2πk,k∈Zx = 2\pi k; \ x = -\frac{\pi}{2} + 2\pi k, \quad k \in \mathbb{Z}x=2πk; x=−2π+2πk,k∈Z