Question

решить тригонометрическое уравнение​

Answer 1

Ответ:

$-\dfrac{\pi }{6} +\dfrac{2\pi k}{3} ,~k\in\mathbb {Z} ; \dfrac{7\pi }{18}+\dfrac{2\pi k}{3},~k\in\mathbb {Z} \end{array} \right.$

Объяснение:

$cos(3x-\dfrac{\pi }{3} )= -\dfrac{\sqrt{3} }{2} ;\\\\3x-\dfrac{\pi }{3}=\pm arccos\left( -\dfrac{\sqrt{3} }{2} \right)+2\pi k,~k\in\mathbb {Z};\\\\3x-\dfrac{\pi }{3}=\pm\dfrac{5\pi }{6} +2\pi k,~k\in\mathbb {Z};\\\\ \left [\begin{array}{l} 3x-\dfrac{\pi }{3}=-\dfrac{5\pi }{6} +2\pi k,~k\in\mathbb {Z} \\ \\3x-\dfrac{\pi }{3}=\dfrac{5\pi }{6} +2\pi k,~k\in\mathbb {Z} \end{array} \right.\Leftrightarrow$

$\left [\begin{array}{l} 3x=-\dfrac{5\pi }{6} +\dfrac{\pi }{3}+2\pi k,~k\in\mathbb {Z} \\ \\3x=\dfrac{5\pi }{6}+\dfrac{\pi }{3} +2\pi k,~k\in\mathbb {Z} \end{array} \right.\Leftrightarrow \left [\begin{array}{l} 3x=-\dfrac{\pi }{2} +2\pi k,~k\in\mathbb {Z} \\ \\3x=\dfrac{7\pi }{6}+2\pi k,~k\in\mathbb {Z} \end{array} \right.\Leftrightarrow$

$\left [\begin{array}{l} x=-\dfrac{\pi }{6} +\dfrac{2\pi k}{3} ,~k\in\mathbb {Z} \\ \\x=\dfrac{7\pi }{18}+\dfrac{2\pi k}{3},~k\in\mathbb {Z} \end{array} \right.$