ГДЗ по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 4

\[\boxed{\mathbf{4.}}\]

\[\textbf{а)}\cos{2x} - cos^{2}x - \sin x = 0\]

\[\text{co}s^{2}x - sin^{2}x -\]

\[- cos^{2}x - \sin x = 0\]

\[- sin^{2}x - \sin x = 0\]

\[\text{si}n^{2}x + \sin x = 0\]

\[\sin x\left( \sin x + 1 \right) = 0\]

\[1)\sin x = 0\]

\[x = \pi k.\]

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

\[\sin x = - 1\]

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

\[\textbf{б)}\ \cos{2x} - cos^{2}x + \sin x = 0\]

\[\text{co}s^{2}x - sin^{2}x - cos^{2}x +\]

\[+ \sin x = 0\]

\[- sin^{2}x + \sin x = 0\]

\[- \sin x\left( \sin x - 1 \right) = 0\]

\[1)\sin x = 0\]

\[x = \pi k.\]

\[2)\sin x - 1 = 0\]

\[\sin x = 1\]

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

\[\textbf{в)}\cos{2x} + cos^{2}x - 0,5 = 0\]

\[2cos^{2}x - 1 + \cos^{2}x - 0,5 = 0\]

\[3\cos^{2}x = 1,5\]

\[\cos^{2}x = 0,5 = \frac{1}{2}\]

\[\cos x = \pm \frac{\sqrt{2}}{2}.\]

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

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

\[2)\cos x = - \frac{\sqrt{2}}{2}\]

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

\[Общее\ решение:\]

\[x = \frac{\pi}{4} + \frac{\text{πk}}{2}.\]

\[\textbf{г)}\cos{2x} - sin^{2}x + 0,5 = 0\]

\[1 - 2sin^{2}x - sin^{2}x + 0,5 = 0\]

\[- 3sin^{2}x = - 1,5\]

\[\text{si}n^{2}x = 0,5 = \frac{1}{2}\]

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

\[\cos{2x} = 0\]

\[2x = \frac{\pi}{2} + \pi k\ \]

\[x = \frac{\pi}{4} + \frac{\text{πk}}{2}.\]