ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 628

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\[1)\ \left( tg\ x - \sqrt{3} \right)\left( 2\sin\frac{x}{12} + 1 \right) = 0\]

\[tg\ x - \sqrt{3} = 0\]

\[tg\ x = \sqrt{3}\]

\[x = arctg\ \sqrt{3} + \pi n\]

\[x = \frac{\pi}{3} + \pi n.\]

\[2\sin\frac{x}{12} + 1 = 0\]

\[2\sin\frac{x}{12} = - 1\]

\[\sin\frac{x}{12} = - \frac{1}{2}\]

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

\[\frac{x}{12} = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]

\[x = 12 \bullet \left( ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n \right)\]

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

\[Ответ:\ \ \frac{\pi}{3} + \pi n;\ \ \]

\[( - 1)^{n + 1} \bullet 2\pi + 12\pi n.\]

\[2)\left( 1 - \sqrt{2}\cos\frac{x}{4} \right)\left( 1 + \sqrt{3}\text{\ tg\ x} \right) = 0\]

\[1 - \sqrt{2}\cos\frac{x}{4} = 0\]

\[\sqrt{2}\cos\frac{x}{4} = 1\]

\[\cos\frac{x}{4} = \frac{1}{\sqrt{2}}\]

\[\frac{x}{4} = \pm \arccos\frac{1}{\sqrt{2}} + 2\pi n\]

\[\frac{x}{4} = \pm \frac{\pi}{4} + 2\pi n\]

\[x = 4 \bullet \left( \pm \frac{\pi}{4} + 2\pi n \right)\]

\[x = \pm \pi + 8\pi n.\]

\[1 + \sqrt{3}\ tg\ x = 0\]

\[\sqrt{3}\ tg\ x = - 1\]

\[tg\ x = - \frac{1}{\sqrt{3}}\]

\[x = - arctg\frac{1}{\sqrt{3}} + \pi n\]

\[x = - \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \pm \pi + 8\pi n;\ \ - \frac{\pi}{6} + \pi n.\]

\[3)\left( 2\sin\left( x + \frac{\pi}{6} \right) - 1 \right)(2\ tg\ x + 1) = 0\]

\[2\sin\left( x + \frac{\pi}{6} \right) - 1 = 0\]

\[2\sin\left( x + \frac{\pi}{6} \right) = 1\]

\[\sin\left( x + \frac{\pi}{6} \right) = \frac{1}{2}\]

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

\[\cos\left( x - \frac{\pi}{3} \right) = \frac{1}{2}\]

\[x - \frac{\pi}{3} = \pm \arccos\frac{1}{2} + 2\pi n\]

\[x - \frac{\pi}{3} = \pm \frac{\pi}{3} + 2\pi n\]

\[x_{1} = - \frac{\pi}{3} + \frac{\pi}{3} + 2\pi n = 2\pi n;\]

\[x_{2} = + \frac{\pi}{3} + \frac{\pi}{3} + 2\pi n =\]

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

\[2\ tg\ x + 1 = 0\]

\[2\ tg\ x = - 1\]

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

\[x = - arctg\frac{1}{2} + \pi n.\]

\[Ответ:\ \ 2\pi n;\ \ \frac{2\pi}{3} + 2\pi n;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } - arctg\frac{1}{2} + \pi n.\]

\[4)\ \left( 1 + \sqrt{2}\cos\left( x + \frac{\pi}{4} \right) \right)(tg\ x - 3) = 0\]

\[1 + \sqrt{2}\cos\left( x + \frac{\pi}{4} \right) = 0\]

\[\sqrt{2}\cos\left( x + \frac{\pi}{4} \right) = - 1\]

\[\cos\left( x + \frac{\pi}{4} \right) = - \frac{1}{\sqrt{2}}\]

\[x + \frac{\pi}{4} = \pm \left( \pi - \arccos\frac{1}{\sqrt{2}} \right) + 2\pi n\]

\[x + \frac{\pi}{4} = \pm \left( \pi - \frac{\pi}{4} \right) + 2\pi n\]

\[x + \frac{\pi}{4} = \pm \frac{3\pi}{4} + 2\pi n\]

\[x_{1} = - \frac{3\pi}{4} - \frac{\pi}{4} + 2\pi n =\]

\[= - \pi + 2\pi n;\]

\[x_{2} = + \frac{3\pi}{4} - \frac{\pi}{4} + 2\pi n =\]

\[= \frac{\pi}{2} + 2\pi n - нет.\]

\[tg\ x - 3 = 0\]

\[tg\ x = 3\]

\[x = arctg\ 3 + \pi n.\]

\[Ответ:\ - \pi + 2\pi n;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ }arctg\ 3 + \pi n.\]