## Giúp mình câu 2b với gấp lắm ạ

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Giúp mình câu 2b với gấp lắm ạ

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8 tháng 2020-10-13T16:05:16+00:00 2 Answers 88 views 0

$x = \left\{\dfrac{53\pi}{84};\dfrac{5\pi}{12};\dfrac{59\pi}{84}\right\}$
$\begin{array}{l} \cos7x – \sqrt3\sin7x = – \sqrt2\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin7x – \dfrac{1}{2}\cos7x = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\left(7x – \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}7x – \dfrac{\pi}{6} = \dfrac{\pi}{4} + k2\pi\\7x – \dfrac{\pi}{6} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}7x =\dfrac{5\pi}{12} + k2\pi\\7x =\dfrac{11\pi}{12} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =\dfrac{5\pi}{84} + k\dfrac{2\pi}{7}\\x =\dfrac{11\pi}{84} + k\dfrac{2\pi}{7}\end{array}\right.\quad (k \in \Bbb Z)\\ Ta\,\,có:\\ \quad \dfrac{2\pi}{5} < x < \dfrac{6\pi}{7}\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{2\pi}{5} <\dfrac{5\pi}{84} + k\dfrac{2\pi}{7} < \dfrac{6\pi}{7}\\\dfrac{2\pi}{5} <\dfrac{11\pi}{84} + k\dfrac{2\pi}{7}< \dfrac{6\pi}{7}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{143}{120}< k < \dfrac{67}{24}\\\dfrac{113}{120} < k< \dfrac{61}{24}\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}k = 2\\k = \left\{1;2\right\}\end{array}\right.\quad (k \in \Bbb Z)\\ \Rightarrow \left[\begin{array}{l}x = \dfrac{29\pi}{84}\\x = \dfrac{5\pi}{12}\\x = \dfrac{59\pi}{84}\end{array}\right.\\ Vậy\,\,x = \left\{\dfrac{53\pi}{84};\dfrac{5\pi}{12};\dfrac{59\pi}{84}\right\} \end{array}$