Страница 193 номер 644, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (620—644). 1) \(4 |\cos x| + 3 = 4 \sin^2 x\); 2) \(|\operatorname{tg} x| + 1 = \frac{1}{\cos^2 x}\).
1) \(4\left| \cos x \right| + 3 = 4\sin^{2}x\)
\(4\left| \cos x \right| + 3 - 4 + 4\cos^{2}x = 0,\)
\(\cos x = z, 4z^{2} + 4|z| - 1 = 0,\)
\(\begin{cases} z \geqslant 0 \\ 4z^{2} + 4z - 1 = 0 \end{cases},\)
\(z_{1{,}2} = \dfrac{-4 \pm \sqrt{16 + 16}}{8},\)
\(z = \dfrac{-4 + 4\sqrt{2}}{8};\)
\(\begin{cases} z < 0 \\ 4z^{2} - 4z - 1 = 0 \end{cases},\)
\(z_{1{,}2} = \dfrac{4 \pm 4\sqrt{2}}{8},\)
\(z = \dfrac{4 - 4\sqrt{2}}{8};\)
\(\cos x = \pm \left( \dfrac{4 - 4\sqrt{2}}{8} \right),\)
\(x_{1} = \pm \arccos\dfrac{\sqrt{2} - 1}{2} + \pi n, n \in Z;\)
2) \(\left| \operatorname{tg} x \right| + 1 = \frac{1}{\cos^{2}x}\)
\(\left| \operatorname{tg} x \right| = \operatorname{tg}^{2}x,\)
\(\left| \operatorname{tg} x \right|\left( \left| \operatorname{tg} x \right| - 1 \right) = 0,\)
\(\operatorname{tg} x = 0\) или \(\operatorname{tg} x = \pm 1\)
\(x_{1} = \pi n,\)
\(x_{2} = \pm \dfrac{\pi}{4} + \pi n, n \in Z.\)
