Question

решите уравнение
$sin4x - sin2x = 0$

Answer 1

$sin4x-sin2x=0$

$2sin2xcos2x-sin2x=0$

$sin2x(2cos2x-1)=0$

$sin2x=0; 2cos2x-1=0$

$x=\frac{\pi k}{2} , k=Z; x=\frac{\pi }{6}+\pi k , k=Z; x=\frac{5\pi }{6} +\pi k , k=Z$

$x=\left \{ {\frac{\pi }{6}+\frac{\pi k}{3}} \atop {\frac{\pi k}{2} }} \right. , k=Z$

Answer 2

$sin(4x) - sin(2x) = 0\\2sin(2x)\cdot cos(2x) - sin(2x) = 0\\\\sin(2x)(2cos2x - 1) = 0$

Произведение равно 0, когда один из множителей равен 0

$1) \;\; sin(2x) = 0\\\\2x = \pi n\\\\x = \frac{\pi}{2} \cdot n,\;\;n \in Z\\\\\\2) \;\; 2cos(2x) - 1 =0\\\\cos(2x) = \frac{1}{2}\\\\2x = \pm arccos(\frac{1}{2}) + 2\pi k, \;\; k \in Z\\\\2x = \pm \frac{\pi}{3} + 2\pi k, \;\; k\in Z\\\\x = \pm \frac{\pi}{6}+ \pi k, \;\; k \in Z$

Ответ: $\frac{\pi}{2} \cdot n, \;\; \frac{\pi}{6}+ \pi k, \;\; -\frac{\pi}{6}+ \pi k$