Question

Решите неравенство: log{2} (sqrt{2} *sinx) =log{4} (cos4x-cos6x)

Answer 1

$log_{2}(\sqrt{2}sinx)=log_{4}(cos4x-cos6x)\\$

ОДЗ:

$\left \{ {{\sqrt{2}sinx0} \atop {cos4x-cos6x0}} \right. \left \{ {{2\pi n < x< \pi +2\pi n, n \in Z} \atop {2sin5xsinx0\Rightarrow sin5x0}} \right. \left \{ {{2\pi n < x< \pi +2\pi n, n \in Z} \atop {{2\pi k$

$x \in (\frac{2}{5}\pi k; \frac{\pi}{5} +\frac{2}{5}\pi k), k \in Z}$

$log_{2}(\sqrt{2}sinx)=log_{2^2}(cos4x-cos6x)$

$log_{2}(\sqrt{2}sinx)=\frac{1}{2} log_{2}(cos4x-cos6x)$

$2\cdot log_{2}(\sqrt{2}sinx)= log_{2}(cos4x-cos6x)$

$log_{2}(\sqrt{2}sinx)^2= log_{2}(cos4x-cos6x)$

По свойству монотонности:

$(\sqrt{2}sinx)^2= (cos4x-cos6x)$

$2sin^2x=2sin5x\cdot sinx$

$2sin^2x-2sin5x\cdot sinx=0$

$2sinx(sinx-sin5x)=0$

$sinx=0$   или    sinx-sin5x=0

$x=\pi n, n\in Z$   или   $2sin3x\cdot cos2x=0$⇒  $x=\frac{\pi}{3}m, m \in Z$  или $x=\frac{\pi}{4}+\frac{\pi}{2} }s, s \in Z$

C учетом ОДЗ получаем ответ:

О т в е т.

$x=\pi n, n\in Z;$

$\frac{\pi}{3}+\pi+2\pi m=\frac{4\pi}{3}+2\pi m, m \in Z$$;\frac{2\pi}{3}+\pi+2\pi m=\frac{5\pi}{3}+2\pi m, m \in Z$

$\frac{\pi}{4}+\pi +2 \pi s=\frac{5\pi}{4} +2 \pi s, s \in Z;$$\frac{3\pi}{4}+\pi +2 \pi s=\frac{7\pi}{4} +2 \pi s, s \in Z.$