Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1375

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\[1)\sin^{3}x + \cos^{3}x = 0\]

\[\left( \sin x + \cos x \right)\left( 1 + 0,5\sin{2x} \right) = 0\]

\[\sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

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

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

\[1 + 0,5\sin x = 0\]

\[0,5\sin x = - 1\]

\[\sin x = - 2\]

\[корней\ нет.\]

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

\[2)\ 2\sin^{2}x + \sin^{2}{2x} = 2\]

\[2\sin^{2}x + 4\sin^{2}x \bullet \cos^{2}x - 2 = 0\]

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

\[2\sin^{2}x + 4\sin^{2}x - 4\sin^{4}x - 2 = 0\]

\[6\sin^{2}x - 4\sin^{4}x - 2 = 0\]

\[2\sin^{4}x - 3\sin^{2}x + 1 = 0\]

\[y = \sin^{2}x:\]

\[2y^{2} - 3y + 1 = 0\]

\[D = 9 - 8 = 1\]

\[y_{1} = \frac{3 - 1}{2 \bullet 2} = \frac{1}{2};\]

\[y_{2} = \frac{3 + 1}{2 \bullet 2} = 1.\]

\[1)\ \sin^{2}x = \frac{1}{2}\]

\[\sin x = \pm \frac{\sqrt{2}}{2}\]

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

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

\[2)\ \sin^{2}x = 1\]

\[\sin x = \pm 1\]

\[x = \pm \arcsin 1 + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2};\ \ \frac{\pi}{2} + \pi n.\]

\[3)\ 8\sin x \bullet \cos{2x} \bullet \cos x = \sqrt{3}\]

\[4\sin{2x} \bullet \cos{2x} = \sqrt{3}\]

\[2\sin{4x} = \sqrt{3}\]

\[\sin{4x} = \frac{\sqrt{3}}{2}\]

\[4x = ( - 1)^{n} \bullet \arcsin\frac{\sqrt{3}}{2} + \pi n\]

\[4x = ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n\]

\[x = \frac{1}{4} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{4}.\]

\[Ответ:\ \ ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{4}.\]

\[4)\ 4\sin x \bullet \cos x \bullet \cos{2x} = \cos{4x}\]

\[2\sin{2x} \bullet \cos{2x} = \cos{4x}\]

\[\sin{4x} = \cos{4x}\ \ \ \ \ |\ :\cos{4x}\]

\[tg\ 4x = 1\]

\[4x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]

\[x = \frac{1}{4} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{16} + \frac{\text{πn}}{4}.\]

\[Ответ:\ \ \frac{\pi}{16} + \frac{\text{πn}}{4}.\]