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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 11: giải pt sau cos bình 3x=3/4 tìm min max hàm số y=4/căn3 sinx+cosx

Toán Lớp 11: giải pt sau
cos bình 3x=3/4
tìm min max hàm số
y=4/căn3 sinx+cosx

$\begin{array}{l} a){\cos ^2}3x = \dfrac{3}{4}\\ \Leftrightarrow \cos 3x = \pm \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l} \cos 3x = \cos \dfrac{\pi }{6}\\ \cos 3x = \cos \dfrac{{5\pi }}{6} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 3x = \pm \dfrac{\pi }{6} + k2\pi \\ 3x = \pm \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \pm \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\ x = \pm \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3} \end{array} \right.\\ Vay\,x = \pm \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3};x = \pm \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}\\ b)y = \dfrac{4}{{\sqrt {3\sin x + \cos x} }}\\ = \dfrac{{\dfrac{4}{{\sqrt {10} }}}}{{\sqrt {\dfrac{3}{{\sqrt {10} }}\sin x + \dfrac{1}{{\sqrt {10} }}\cos x} }}\\ = \dfrac{4}{{\sqrt {10} }}.\dfrac{1}{{\sqrt {\sin \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right)} }}\\ Do:0 < \sqrt {\sin \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right)} \le 1\\ \Leftrightarrow \dfrac{1}{{\sqrt {\sin \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right)} }} \ge 1\\ \Leftrightarrow \dfrac{4}{{\sqrt {10} }}.\dfrac{1}{{\sqrt {\sin \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right)} }} \ge \dfrac{4}{{\sqrt {10} }}\\ \Leftrightarrow y \ge \dfrac{{2\sqrt {10} }}{5}\\ \Leftrightarrow \min y = \dfrac{{2\sqrt {10} }}{5}\,\\ Khi:\sin \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right) = 1\\ \Leftrightarrow \left( {x + \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right)} \right) = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{2} – \arcsin \left( {\dfrac{3}{{\sqrt {10} }}} \right) + k2\pi \end{array}$