Question

Sin4x= cos (22pi- 2x)

Answer 1

Ответ:

$\dfrac{\pi }{4} +\dfrac{\pi k}{2} ,~k\in\mathbb {Z} ; \dfrac{\pi }{12} +\pi n,~n\in\mathbb {Z} ; \dfrac{5\pi }{12} +\pi m,~m\in\mathbb {Z}.$

Объяснение:

Решим уравнение

$sin4x= cos(22\pi -2x);\\sin4x=cos2x .$

Применим формулу синуса двойного угла

$sin2x=2\cdot sinx\cdot cosx$

$2sin2x\cdot cos2x =cos2x ;\\2sin2x\cdot cos2x -cos2x =0;\\cos2x(2sin2x-1)=0;$

$\left [\begin{array}{l} cos2x= 0, \\ 2sin2x-1 = 0; \end{array} \right.\Leftrightarrow \left [\begin{array}{l} cos2x= 0, \\ sin2x= \dfrac{1}{2} ; \end{array} \right.\Leftrightarrow \left [\begin{array}{l} 2x = \dfrac{\pi }{2} +\pi k,~k\in\mathbb {Z} \\ \\2x = \dfrac{\pi }{6} +2\pi n,~n\in\mathbb {Z} \\\\ 2x = \dfrac{5\pi }{6} +2\pi m,~m\in\mathbb {Z}\end{array} \right.\Leftrightarrow$

$\right.\Leftrightarrow \left [\begin{array}{l} x = \dfrac{\pi }{4} +\dfrac{\pi k}{2} ,~k\in\mathbb {Z} \\ \\x = \dfrac{\pi }{12} +\pi n,~n\in\mathbb {Z} \\\\ x = \dfrac{5\pi }{12} +\pi m,~m\in\mathbb {Z}\end{array} \right.$