Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 638

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\[1)\sin^{2}x + \sin^{2}{2x} = \sin^{2}{3x}\]

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

\[\left( \sin x - \sin{3x} \right)\left( \sin x + \sin{3x} \right) + \sin^{2}{2x} = 0\]

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

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

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

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

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

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

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

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

\[2x = \pm \arccos\frac{1}{2} + 2\pi n\]

\[2x = \pm \frac{\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right)\]

\[x = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \frac{\text{πn}}{2};\ \ \pm \frac{\pi}{6} + \pi n.\]

\[y = \sin x + \cos x:\]

\[y + \frac{y^{2} - 1}{2} \bullet y = 2y^{2}\]

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

\[2y + y^{3} - y - 4y^{2} = 0\]

\[y^{3} - 4y^{2} + y = 0\]

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

\[D = 4^{2} - 4 = 16 - 4 = 12\]

\[y = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} =\]

\[= 2 \pm \sqrt{3}.\]

\[1)\ \sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

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

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

\[2)\ \sin x + \cos x = 2 - \sqrt{3}\ \ \ \ \ |\ :\sqrt{2}\]

\[\frac{\sqrt{2}}{2} \bullet \sin x + \cos x \bullet \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{3}}{\sqrt{2}}\]

\[\cos\frac{\pi}{4} \bullet \sin x + \cos x \bullet \sin\frac{\pi}{4} = \frac{2 - \sqrt{3}}{\sqrt{2}}\]

\[\sin\left( x + \frac{\pi}{4} \right) = \frac{2 - \sqrt{3}}{\sqrt{2}}\]

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

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

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