ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 680

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

\[2\cos{3x} + 2\cos x = 3\sin x + 3\cos x\]

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

\[= 3\left( \sin x + \cos x \right)\]

\[4 \bullet \cos{2x} \bullet \cos x = 3\left( \sin x + \cos x \right)\]

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

\[= 3\left( \sin x + \cos x \right)\]

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

\[y = tg\ x:\]

\[1 - 4y - 3y^{2} = 0\]

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

\[D = 16 + 12 = 28 = 4 \bullet 7\]

\[y = \frac{- 4 \pm 2\sqrt{7}}{2 \bullet 3} = \frac{- 2 \pm \sqrt{7}}{3}.\]

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

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

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

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

\[2)\ tg\ x = \frac{- 2 - \sqrt{7}}{3} = - \frac{2 + \sqrt{7}}{3}\]

\[x = - arctg\frac{2 + \sqrt{7}}{3} + \pi n.\]

\[3)\ tg\ x = \frac{- 2 + \sqrt{7}}{3}\]

\[x = arctg\frac{\sqrt{7} - 2}{3} + \pi n.\]

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

\[- arctg\frac{2 + \sqrt{7}}{3} + \pi n;\ \ \]

\[\text{arctg}\frac{\sqrt{7} - 2}{3} + \pi n.\]

\[2)\cos{3x} - \cos{2x} = \sin{3x}\]

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

\[4 + 4 \bullet \frac{y^{2} - 1}{2} - 3 - y = 0\]

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

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

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2 \bullet 2} = - \frac{2}{4} = - \frac{1}{2};\ \]

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

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

\[1 + tg\ x = 0\]

\[tg\ x = - 1\]

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

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

\[2)\ \cos x - \sin x = - \frac{1}{2}\ \ \ \ \ |\ :\sqrt{2}\]

\[\frac{\sqrt{2}}{2} \bullet \cos x - \frac{\sqrt{2}}{2} \bullet \sin x = - \frac{1}{2\sqrt{2}}\]

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

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

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

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

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

\[3)\ \cos x - \sin x = 1\ \ \ \ \ |\ :\ \sqrt{2}\]

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

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

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

\[x + \frac{\pi}{4} = \pm \arccos\frac{\sqrt{2}}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{4} + 2\pi n;\]

\[x_{1} = - \frac{\pi}{4} - \frac{\pi}{4} + 2\pi n =\]

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

\[x_{2} = + \frac{\pi}{4} - \frac{\pi}{4} + 2\pi n = 2\pi n.\]

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

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

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