Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1168

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\[1)\ 1 - 4\sin x \bullet \cos x = 0\]

\[2 \bullet 2\sin x \bullet \cos x = 1\]

\[\sin{2x} = \frac{1}{2}\]

\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n =\]

\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}\]

\[Ответ:\ \ x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]

\[2)\ \sqrt{3} + 4\sin x \bullet \cos x = 0\]

\[2 \bullet 2\sin x \bullet \cos x = - \sqrt{3}\]

\[\sin{2x} = - \frac{\sqrt{3}}{2}\]

\[2x = ( - 1)^{n + 1} \bullet \arcsin\frac{\sqrt{3}}{2} + \pi n =\]

\[= ( - 1)^{n + 1} \bullet \frac{\pi}{3} + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n + 1} \bullet \frac{\pi}{3} + \pi n \right)\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \frac{\text{πn}}{2}\]

\[Ответ:\ \ x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \frac{\text{πn}}{2}.\]

\[3)\ 1 + 6\sin\frac{x}{4} \bullet \cos\frac{x}{4} = 0\]

\[3 \bullet 2\sin\frac{x}{4} \bullet \cos\frac{x}{4} = - 1\]

\[\sin\frac{x}{2} = - \frac{1}{3}\]

\[\frac{x}{2} = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{3} + \pi n\]

\[x = ( - 1)^{n + 1} \bullet 2\arcsin\frac{1}{3} + 2\pi n\]

\[Ответ:\ \ x = ( - 1)^{n + 1} \bullet\]

\[\bullet 2\arcsin\frac{1}{3} + 2\pi n.\]

\[4)\ 1 - 8\sin\frac{x}{3} \bullet \cos\frac{x}{3} = 0\]

\[4 \bullet 2\sin\frac{x}{3} \bullet \cos\frac{x}{3} = 1\]

\[\sin\frac{2x}{3} = \frac{1}{4}\]

\[\frac{2x}{3} = ( - 1)^{n} \bullet \arcsin\frac{1}{4} + \pi n\]

\[x = ( - 1)^{n}\ \bullet \frac{3}{2}\arcsin\frac{1}{4} + \frac{3\pi n}{2}\]

\[Ответ:\ x = \ ( - 1)^{n}\ \bullet \frac{3}{2}\arcsin\frac{1}{4} +\]

\[+ \frac{3\pi n}{2}\text{.\ \ }\]