Tìm x trên đoạn [ -3pi , 3pi] Biết a) sinx =1/2 b) sinx =1

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1. a) $\sin x = \dfrac{1}{2}$

   $\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

   Ta có: $x \in [-3\pi; 3\pi]$

   $\Leftrightarrow \left[\begin{array}{l}-3\pi \leq \dfrac{\pi}{6} + k2\pi \leq 3\pi\\-3\pi \dfrac{5\pi}{6} + k2\pi \leq 3\pi\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}-\dfrac{19}{12}\leq k \leq \dfrac{17}{12}\\-\dfrac{23}{12} \leq k \leq \dfrac{13}{12}\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}k = \left\{-1;0;1\right\}\\k = \left\{-1;0;1\right\}\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{11\pi}{6}\\x = \dfrac{\pi}{6}\\x = \dfrac{13\pi}{6}\\x =- \dfrac{7\pi}{6}\\x = \dfrac{5\pi}{6}\\x = \dfrac{17\pi}{6}\end{array}\right.$

   b) $\sin x = 1$

   $\Leftrightarrow x  = \dfrac{\pi}{2} + k2\pi$

   Ta có: $x \in [-3\pi;3\pi]$

   $\Leftrightarrow – 3\pi \leq x  = \dfrac{\pi}{2} + k2\pi \leq 3\pi$

   $\Leftrightarrow -\dfrac{7}{4} \leq k \leq \dfrac{5}{4}$

   $\Rightarrow k = \left\{-1;0;1\right\}$

   $\Rightarrow \left[\begin{array}{l} x = -\dfrac{3\pi}{2}\\x = \dfrac{\pi}{2}\\x = \dfrac{5\pi}{2}\end{array}\right.$

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