\( 16^{\sin x} = \left( \dfrac{1}{4} \right)^{2\sin 2x} \)

\( 16 = 4^2 \), \( \dfrac{1}{4} = 4^{-1} \)

\[(4^2)^{\sin x} = (4^{-1})^{2\sin 2x}\]

\[4^{2\sin x} = 4^{-2\sin 2x}\]

Показатели равны:

\[2\sin x = -2\sin 2x\]

\[\sin x = -\sin 2x\]

\[\sin 2x = 2\sin x \cos x\]

\[\sin x = -2\sin x \cos x\]

\[\sin x + 2\sin x \cos x = 0\]

\[\sin x (1 + 2\cos x) = 0\]

\[\left[\begin{array}{l}\sin x = 0 \\1 + 2\cos x = 0\end{array}\right.\]

\[x = \pi k,\quad k \in \mathbb{Z}\]

\[2\cos x = -1 \implies \cos x = -\dfrac{1}{2}\]

\[x = \pm \dfrac{2\pi}{3} + 2\pi k,\quad k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z}\]

\( x \in \left[2\pi;\; \dfrac{7\pi}{2} \right] \)

Подставляем значения \(k\) и выбираем подходящие:

\[\begin{array}{ll}x = \pi k: & k = 2 \implies x = 2\pi \\ & k = 3 \implies x = 3\pi \\x = \dfrac{2\pi}{3} + 2\pi k: & k = 1 \implies x = \dfrac{8\pi}{3} \\x = -\dfrac{2\pi}{3} + 2\pi k: & k = 2 \implies x = \dfrac{10\pi}{3}\end{array}\]

\[\left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z}\]

\[x = 2\pi;\quad x = \dfrac{8\pi}{3};\quad x = 3\pi;\quad x = \dfrac{10\pi}{3}\]

\( \begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б)} & x = 2\pi;\; \dfrac{8\pi}{3};\; 3\pi;\; \dfrac{10\pi}{3} \quad \text{✅}\end{array}\)

\( \begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б)} & x = 2\pi;\; \dfrac{7\pi}{3};\; 3\pi;\; \dfrac{11\pi}{3}\end{array}\)

\( \begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б)} & x = 2\pi;\; \dfrac{7\pi}{3};\; 3\pi;\; \dfrac{10\pi}{3}\end{array}\)

\( \begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б)} & x = 2\pi;\; \dfrac{8\pi}{3};\; 3\pi;\; \dfrac{11\pi}{3}\end{array}\)

\( \begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б)} & x = 2\pi;\; \dfrac{8\pi}{3};\; 3\pi;\; \dfrac{9\pi}{3}\end{array}\)

\[\left(25^{\cos x}\right)^{\sin x} = 5^{\cos x}\]

\[25^{\cos x \cdot \sin x} = 5^{\cos x}\]

\[(5^2)^{\cos x \cdot \sin x} = 5^{\cos x}\]

\[5^{2\cos x \cdot \sin x} = 5^{\cos x}\]

\[2\cos x \cdot \sin x = \cos x\]

\[2\cos x \cdot \sin x — \cos x = 0\]

\[\cos x (2\sin x — 1) = 0\]

\[\left[\begin{array}{l}\cos x = 0 \\2\sin x — 1 = 0\end{array}\right.\]

\[\cos x = 0 \implies x = \dfrac{\pi}{2} + \pi k,\quad k \in \mathbb{Z}\]

\[2\sin x — 1 = 0 \implies \sin x = \dfrac{1}{2}\]

\[x = \dfrac{\pi}{6} + 2\pi k,\quad x = \dfrac{5\pi}{6} + 2\pi k,\quad k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \dfrac{\pi}{2} + \pi k \\x = \dfrac{\pi}{6} + 2\pi k \\x = \dfrac{5\pi}{6} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

k = -3:\quad x = \dfrac{\pi}{2} — 3\pi = -\dfrac{5\pi}{2} \\

k = -2:\quad x = \dfrac{\pi}{2} — 2\pi = -\dfrac{3\pi}{2}

k = -1:\quad x = \dfrac{\pi}{6} — 2\pi = -\dfrac{11\pi}{6}

k = -1:\quad x = \dfrac{5\pi}{6} — 2\pi = -\dfrac{7\pi}{6}

\[\begin{array}{ll}\text{1.} & x = -\dfrac{5\pi}{2},\; -\dfrac{11\pi}{6},\; -\dfrac{3\pi}{2},\; -\dfrac{7\pi}{6} \;\; \text{✅} \\\text{2.} & x = -\dfrac{5\pi}{2},\; -\dfrac{3\pi}{2},\; -\dfrac{7\pi}{6} \\\text{3.} & x = -\dfrac{3\pi}{2},\; -\dfrac{11\pi}{6},\; -\dfrac{7\pi}{6} \\\text{4.} & x = -\dfrac{5\pi}{2},\; -\dfrac{11\pi}{6},\; -\dfrac{7\pi}{6} \\\text{5.} & x = -\dfrac{5\pi}{2},\; -\dfrac{3\pi}{2},\; -\dfrac{11\pi}{6}\end{array}\]

\[36^{\sin 2x} = 6^{2\sin x}\]

\[36 = 6^2 \implies 36^{\sin 2x} = (6^2)^{\sin 2x} = 6^{2\sin 2x}\]

Уравнение принимает вид:

\[6^{2\sin 2x} = 6^{2\sin x}\]

\[2\sin 2x = 2\sin x \implies \sin 2x = \sin x\]

\[\sin 2x = 2\sin x \cos x\]

Подставляем:

\[2\sin x \cos x = \sin x\]

\[2\sin x \cos x — \sin x = 0\]

\[\sin x (2\cos x — 1) = 0\]

\[\left[\begin{array}{l}\sin x = 0 \\2\cos x — 1 = 0\end{array}\right.\]

Решения:

\[\sin x = 0 \implies x = \pi k,\quad k \in \mathbb{Z}\]

\[2\cos x — 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \pm\frac{\pi}{3} + 2\pi k,\quad k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \pi k \\x = \pm\dfrac{\pi}{3} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\[\left[ -\dfrac{7\pi}{2};\ -\dfrac{5\pi}{2} \right]\]

Проверяем решения из совокупности:

Для \( x = \pi k \):

\[x = -3\pi \quad (k = -3)\]

\[-3\pi \in \left[ -\dfrac{7\pi}{2};\ -\dfrac{5\pi}{2} \right]\]

Другие значения не попадают в промежуток.

Для \( x = \pm\dfrac{\pi}{3} + 2\pi k \) подходящих значений нет.

\[\left[\begin{array}{l}x = \pi k \\x = \pm\dfrac{\pi}{3} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\( x = -3\pi \) ✅

\( x = -2\pi \)

\( x = -\dfrac{11\pi}{3} \)

\( x = -\dfrac{7\pi}{2} \)

\( x = -\dfrac{5\pi}{2} \)

\[\frac{1}{25} = 5^{-2}\]

Подставим:

\[5^{2\sin 2x} = (5^{-2})^{\cos\left( \frac{3\pi}{2} + x \right)}\]

\[(5^{-2})^{\cos\left( \frac{3\pi}{2} + x \right)} = 5^{-2\cos\left( \frac{3\pi}{2} + x \right)}\]

Теперь уравнение:

\[5^{2\sin 2x} = 5^{-2\cos\left( \frac{3\pi}{2} + x \right)}\]

\[2\sin 2x = -2\cos\left( \frac{3\pi}{2} + x \right)\]

Разделим обе части на 2:

\[\sin 2x = -\cos\left( \frac{3\pi}{2} + x \right)\]

\[\cos\left( \frac{3\pi}{2} + x \right) = \sin x\]

Тогда:

\[\sin 2x = -\sin x\]

\[\sin 2x = 2\sin x \cos x\]

Подставим:

\[2\sin x \cos x = -\sin x\]

\[2\sin x \cos x + \sin x = 0\]

Вынесем \(\sin x\) за скобку:

\[\sin x (2\cos x + 1) = 0\]

Рассматриваем совокупность:

\[x = \pi k, \quad k \in \mathbb{Z}\]

\[\cos x = -\frac{1}{2}\]

\[x = \pm \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}\]

\[\left[\begin{array}{lr}x = \pi k \\x = \pm \dfrac{2\pi}{3} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\[\begin{array}{l}\text{а) } x = \pi k;\ x = \pm \dfrac{2\pi}{3} + 2\pi k,\ k \in \mathbb{Z} \\\text{б) } x = 2\pi;\ \dfrac{8\pi}{3};\ 3\pi\end{array}\]

\[15^{\sin x} = (5 \cdot 3)^{\sin x} = 5^{\sin x} \cdot 3^{\sin x}\]

Уравнение примет вид:

\[5^{\sin x} \cdot 3^{\sin x} = 5^{\sin x} \cdot 3^{ — \sqrt{3} \cos x}\]

\[5^{\sin x} \ne 0\]

Делим обе части на \(5^{\sin x}\):

\[3^{\sin x} = 3^{ — \sqrt{3} \cos x}\]

\[\sin x = -\sqrt{3} \cos x\]

\[\tan x = -\sqrt{3}\]

\[x = -\dfrac{\pi}{3} + \pi k, \quad k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = -\dfrac{\pi}{3} + \pi k\end{array}\right.\quad k \in \mathbb{Z}\]

Промежуток: \(\left[\dfrac{5\pi}{2};\;4\pi\right]\)

Подставляем \(k = 3\):

\[x = -\dfrac{\pi}{3} + 3\pi = \dfrac{8\pi}{3}\]

Подставляем \(k = 4\):

\[x = -\dfrac{\pi}{3} + 4\pi = \dfrac{11\pi}{3}\]

Оба значения принадлежат отрезку.

\[\begin{array}{ll}\text{1.}\quad x = \dfrac{7\pi}{3};\; x = \dfrac{10\pi}{3} & \\\text{2.}\quad x = \dfrac{8\pi}{3};\; x = \dfrac{11\pi}{3} & \quad \text{✅} \\\text{3.}\quad x = \dfrac{5\pi}{3};\; x = \dfrac{8\pi}{3} & \\\text{4.}\quad x = \dfrac{11\pi}{3};\; x = \dfrac{14\pi}{3} & \\\text{5.}\quad x = \dfrac{7\pi}{3};\; x = \dfrac{11\pi}{3} &\end{array}\]