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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 865
Задание 865
\[1)\sin{2x} = 3\cos x\]
\[2\sin x \bullet \cos x - 3\cos x = 0\]
\[\cos x \bullet \left( 2\sin x - 3 \right) = 0\]
\[1)\ \cos x = 0\]
\[x = \frac{\pi}{2} + \pi n.\]
\[2)\ 2\sin x - 3 = 0\]
\[2\sin x = 3\]
\[\sin x = \frac{3}{2}\]
\[x \in \varnothing.\]
\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]
\[2)\sin{4x} = \cos^{4}x - \sin^{4}x\]
\[\sin{4x} =\]
\[= \left( \cos^{2}x - \sin^{2}x \right)\left( \sin^{2}x + \cos^{2}x \right)\]
\[2\sin{2x} \bullet \cos{2x} = \cos{2x} \bullet 1\]
\[\cos{2x} \bullet \left( 2\sin{2x} - 1 \right) = 0\]
\[1)\ \cos{2x} = 0\]
\[2x = \frac{\pi}{2} + \pi n\]
\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]
\[2)\ 2\sin{2x} - 1 = 0\]
\[2\sin{2x} = 1\]
\[\sin{2x} = \frac{1}{2}\]
\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n =\]
\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]
\[x = \frac{1}{2} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right) =\]
\[= ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]
\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2};\ \]
\[( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]
\[3)\ 2\cos^{2}x = 1 + 4\sin{2x}\]
\[2\cos^{2}x - \left( \cos^{2}x + \sin^{2}x \right) = 4\sin{2x}\]
\[\cos^{2}x - \sin^{2}x = 4\sin{2x}\]
\[\cos{2x} = 4\sin{2x}\ \ \ \ \ |\ :\sin{2x}\]
\[ctg\ 2x = 4\]
\[2x = arcctg\ 4 + \pi n\]
\[x = \frac{1}{2} \bullet (arcctg\ 4 + \pi n) =\]
\[= \frac{1}{2}arcctg\ 4 + \frac{\text{πn}}{2}.\]
\[Ответ:\ \ \frac{1}{2}arcctg\ 4 + \frac{\text{πn}}{2}.\]
\[4)\ 2\cos x + \cos{2x} = 2\sin x\]
\[2\cos x - 2\sin x + \cos{2x} = 0\]
\[2\left( \cos x - \sin x \right) + \left( \cos^{2}x - \sin^{2}x \right) = 0\]
\[\left( \cos x - \sin x \right)\left( 2 + \cos x + \sin x \right) = 0\]
\[1)\ \cos x - \sin x = 0\ \ \ \ \ |\ :\cos x\]
\[1 - tg\ x = 0\]
\[tg\ x = 1\]
\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]
\[2)\ 2 + \cos x + \sin x = 0\]
\[\cos x + \sin x = - 2\]
\[x \in \varnothing.\]
\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]
