Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 10

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\[\textbf{а)}\ 1 + \cos x = |\sin x|\]

\[- 1 \leq \cos x \leq 1\]

\[0 \leq 1 + \cos x \leq 2\]

\[x \in R.\]

\[\left( 1 + \cos x \right)^{2} = sin^{2}x\]

\[1 + 2\cos x + cos^{2}x = sin^{2}x\]

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

\[2\cos x\left( \cos x + 1 \right) = 0\]

\[1)\ 2\cos x = 0\]

\[\cos x = 0\]

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

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

\[\cos x = - 1\]

\[x = \pi + 2\pi m.\]

\[Ответ:x = \frac{\pi}{2} + \pi k;x = \pi + 2\pi m.\]

\[\textbf{б)}\ 1 - \sin x = |\cos x|\]

\[- 1 \leq \sin x \leq 1\]

\[0 \leq 1 - \sin x \leq 2\]

\[x \in R.\]

\[\left( 1 - \sin x \right)^{2} = cos^{2}x\]

\[1 - 2\sin x + sin^{2}x = cos^{2}x\]

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

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

\[2\sin x\left( \sin x - 1 \right) = 0\]

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

\[\sin x = 0\]

\[x = \pi k.\]

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

\[\sin x = 1\]

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

\[Ответ:x = \pi k;x = \frac{\pi}{2} + 2\pi m.\]

\[\textbf{в)}\ \sqrt{1 - \cos x} = |\sin x|\]

\[- 1 \leq \cos x \leq 1\]

\[0 \leq 1 - \cos x \leq 2\]

\[x \in R;\]

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

\[1 - sin^{2}x - \cos x = 0\]

\[\text{co}s^{2}x - \cos x = 0\]

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

\[{1)\ cos}x = 0\]

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

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

\[\cos x = 1\]

\[x = 2\pi m.\]

\[Ответ:x = \frac{\pi}{2} + \pi k;x = 2\pi m.\]

\[\textbf{г)}\ \sqrt{1 + \sin x} = \left| \cos x \right|\]

\[- 1 \leq \sin x \leq 1\]

\[0 \leq 1 + \sin x \leq 2\]

\[x \in R.\]

\[1 + \sin x = cos^{2}x\]

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

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

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

\[{1)\ sin}x = 0\]

\[x = \pi k.\]

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

\[\sin x = - 1\]

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

\[Ответ:x = \pi k;x = - \frac{\pi}{2} + 2\pi m.\]