Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 629

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\[1)\ \sqrt{3}\sin x \bullet \cos x = \sin^{2}x\]

\[\sqrt{3}\sin x \bullet \cos x - \sin^{2}x = 0\]

\[\sin x \bullet \left( \sqrt{3}\cos x - \sin x \right) = 0\]

\[\sin x = 0\]

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

\[\sqrt{3}\cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[\sqrt{3} - tg\ x = 0\]

\[tg\ x = \sqrt{3}\]

\[x = arctg\ \sqrt{3} + \pi n\]

\[x = \frac{\pi}{3} + gn.\]

\[Ответ:\ \ \pi n;\ \ \frac{\pi}{3} + \pi n.\]

\[2)\ 2\sin x \bullet \cos x = \cos x\]

\[2\sin x \bullet \cos x - \cos x = 0\]

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

\[\cos x = 0\]

\[x = arctg\ 0 + \pi n = \frac{\pi}{2} + \pi n.\]

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

\[2\sin x = 1\]

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

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

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

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

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

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

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

\[2x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{2} \bullet \pi n = \frac{\text{πn}}{2}.\]

\[2\cos{2x} + \sin{2x} = 0\text{\ \ }\ \ |\ :\cos{2x}\]

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

\[tg\ 2x = - 2\]

\[2x = - arctg\ 2 + \pi n\]

\[x = \frac{1}{2} \bullet ( - arctg\ 2 + \pi n)\]

\[x = - \frac{1}{2}arctg\ 2 + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\text{πn}}{2};\ \ - \frac{1}{2}arctg\ 2 + \frac{\text{πn}}{2}.\]

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

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

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

\[2\cos x = 0\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n\]

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

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

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

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

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

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