ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 644

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\[1)\ 4 \bullet \left| \cos x \right| + 3 = 4\sin^{2}x\]

\[4 \bullet \left| \cos x \right| + 3 - 4\left( 1 - \cos^{2}x \right) = 0\]

\[4 \bullet \left| \cos x \right| + 3 - 4 + 4\cos^{2}x = 0\]

\[4\cos^{2}x + 4 \bullet \left| \cos x \right| - 1 = 0\]

\[y = \cos x:\]

\[4y^{2} + 4|y| - 1 = 0\]

\[4y^{2} \pm 4y - 1 = 0\]

\[D = 16 + 16 = 16 \bullet 2\]

\[y = \frac{\pm 4 \pm 4\sqrt{2}}{2 \bullet 4} = \frac{\pm 4 \pm 4\sqrt{2}}{8} =\]

\[= \frac{\pm 1 \pm \sqrt{2}}{2},\]

\[\lbrack - 1\ 1\rbrack:\]

\[y_{1} = \frac{- 1 + \sqrt{2}}{2};y_{2} = \frac{1 - \sqrt{2}}{2};\]

\[y_{2} = - y_{1}.\]

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

\[x_{1} = \pm \arccos\frac{1 - \sqrt{2}}{2} + 2\pi n;\]

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

\[Ответ:\ \ \pm \arccos\frac{1 - \sqrt{2}}{2} + \pi n.\]

\[2)\ \left| \text{tg\ x} \right| + 1 = \frac{1}{\cos^{2}x}\]

\[\left| \text{tg\ x} \right| = \frac{1}{\cos^{2}x} - 1\]

\[\left| \text{tg\ x} \right| = tg^{2}\text{\ x}\]

\[y = tg\ x:\]

\[|y| = y^{2}\]

\[y^{2} \pm y = 0\]

\[y \bullet (y \pm 1) = 0\]

\[y_{1} = \pm 1;\ y_{2} = 0.\]

\[1)\ tg\ x = \pm 1\]

\[x = \pm arctg\ 1 + \pi n\]

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

\[2)\ tg\ x = 0\]

\[x = arctg\ 0 + \pi n = \pi n.\]

\[Ответ:\ \ \pi n;\ \ \pm \frac{\pi}{4} + \pi n.\]