3] 2sin x-1=0...

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Решите уравнение и найдите его корни на [-5п/2; п/3] 2sin x-1=0

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NNNLLL54

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Ответ:

$2sinx-1=0\\\\sinx=\dfrac{1}{2}\ \ \ \to \ \ \ \left[\begin{array}{l}\ x_1=\dfrac{\pi }{6}+2\pi n\ ,\ n\in Z\\x_2=\dfrac{5\pi }{6}+2\pi k\ ,\ k\in Z\end{array}\right\\\\\\x\in \Big[-\dfrac{5\pi}{2}\ ;\ \dfrac{\pi}{3}\ \Big]:\ x=-\dfrac{11\pi }{6}\ ,\ -\dfrac{7\pi}{6}\ ,\ \dfrac{\pi }{6}\ .$

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