Уравнения

Упростим уравнение: $$2\sin 2x - 2\sin x - 2\cos x + 1 = 0$$ $$4\sin x \cos x - 2\sin x - 2\cos x + 1 = 0$$ $$2\sin x (2\cos x - 1) - (2\cos x - 1) = 0$$ $$(2\cos x - 1)(2\sin x - 1) = 0 \Leftrightarrow \left[ \begin{aligned} &\cos x = \dfrac{1}{2} \\ &\sin x = \dfrac{1}{2} \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \pm\dfrac{\pi}{3} + 2\pi k \\ &x = \dfrac{\pi}{6} + 2\pi k \\ &x = \dfrac{5\pi}{6} + 2\pi k \end{aligned} \right., \quad k \in \mathbb{Z}$$

б) Отберём корни на отрезке $\left[ -4\pi; -\frac{5\pi}{2} \right]$ при помощи тригонометрической окружности:

$$\frac{5\pi}{6} - 4\pi = -\frac{19\pi}{6} \quad$$ $$\frac{\pi}{6} - 4\pi = -\frac{23\pi}{6} \quad$$ $$\frac{\pi}{3} - 4\pi = -\frac{11\pi}{3} \quad$$

Корни, принадлежащие заданному отрезку: $$-\frac{23\pi}{6}; -\frac{11\pi}{3};-\frac{19\pi}{6}$$

Ответ: а) $\pm \frac{\pi}{3} + 2\pi k; \frac{\pi}{6} + 2\pi k; \frac{5\pi}{6} + 2\pi k, \quad k \in \mathbb{Z};$ б) $-\frac{23\pi}{6}; -\frac{11\pi}{3};-\frac{19\pi}{6}$

$$\sin x (\cos 2x + 1) = \sqrt{3} \cos^2 x$$ $$\sin x (2\cos^2 x - 1 + 1) = \sqrt{3} \cos^2 x$$ $$2\sin x \cos^2 x = \sqrt{3} \cos^2 x$$ $$\cos^2 x (2\sin x - \sqrt{3}) = 0 \Leftrightarrow \left[ \begin{aligned} &\cos x = 0 \\ &\sin x = \dfrac{\sqrt3}{2} \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \dfrac{\pi}{2} + \pi k \\ &x = \dfrac{\pi}{3} + 2\pi k \\ &x = \dfrac{2\pi}{3} + 2\pi k \end{aligned} \right., \quad k \in \mathbb{Z}$$

б) Отберём корни на отрезке $\left[ \frac{7\pi}{2}; \frac{9\pi}{2} \right]$ при помощи тригонометрической окружности:

$$\frac{\pi}{3} + 4\pi = \frac{13\pi}{3}$$

Корни, принадлежащие отрезку: $$\frac{7\pi}{2}; \frac{13\pi}{3}; \frac{9\pi}{2}$$

Ответ: а) $\frac{\pi}{2} + \pi k; \frac{\pi}{3} + 2\pi k; \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z};$ б) $\frac{7\pi}{2}; \frac{13\pi}{3}; \frac{9\pi}{2}$

$$4\sin^3x + 4\sqrt{3}(1 - \sin^2x) + 3\sin x - 4\sqrt{3} = 0$$ $$4\sin^3x + 4\sqrt{3} - 4\sqrt{3}\sin^2x + 3\sin x - 4\sqrt{3} = 0$$ $$4\sin^3x - 4\sqrt{3}\sin^2x + 3\sin x = 0$$ $$\sin x(4\sin^2x - 4\sqrt{3}\sin x + 3) = 0 \Leftrightarrow \left[ \begin{aligned} &\sin x = 0 \\ &4\sin^2x - 4\sqrt{3}\sin x + 3 = 0 \end{aligned} \right. \Leftrightarrow$$ $$\Leftrightarrow \left[ \begin{aligned} &x = \pi k, k \in Z \\ & (2\sin x - \sqrt3)^2 = 0 \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \pi k, k \in Z \\ & \sin x = \frac{\sqrt3}{2} \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \pi k\\ &x = \frac{\pi}{3} + 2\pi k \\ & x = \frac{2\pi}{3} + 2\pi k \end{aligned} \right., k \in Z$$

Ответ: $x = \pi k; \ x = \frac{\pi}{3} + 2\pi k; \ x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$

$$2\sin x \cos x - 2\sin x + 2\cos x - 2 = 0$$ $$2\sin x (\cos x - 1) + 2(\cos x - 1) = 0$$ $$2(\cos x - 1)(\sin x + 1) = 0 \Leftrightarrow (\cos x - 1)(\sin x + 1) \Leftrightarrow$$ $$\Leftrightarrow \left[ \begin{aligned} &\sin x = -1 \\ &\cos x = 1 \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} & x = -\frac{\pi}{2} + 2\pi k\\ & x = 2\pi k \end{aligned} \right., k\in Z$$

Ответ: $x = 2\pi k; \ x = -\frac{\pi}{2} + 2\pi k, \quad k \in \mathbb{Z}$

а)

По свойству логарифмов возведём основание и аргумент правого логарифма в степень $\frac{1}{2}$:

$$\log_7(x + 2) = \log_{49}(x^4) \Leftrightarrow \log_7(x + 2) = \log_7(x^2) \Leftrightarrow$$ $$\Leftrightarrow \begin{cases} x + 2 = x^2 \\ x^2 > 0 \end{cases} \Leftrightarrow \begin{cases} x^2 - x - 2 = 0 \\ x \neq 0 \end{cases} \Leftrightarrow \left[ \begin{aligned} &x = -1 \\ &x = 2 \end{aligned} \right.$$

б) Сравним числа:

$$\log_6\frac{1}{7} \leq -1 \leq \log_635 \Leftrightarrow -\log_67 \leq -\log_66 \leq \log_635 \text{ верно}$$ $$\log_6\frac{1}{7} \leq 2 \leq \log_635 \Leftrightarrow \log_6\frac{1}{7} \leq \log_636 \leq \log_635 \text{ неверно}$$

Ответ: a)$-1; 2$; б) $-1$;