Toán Lớp 11: giải pt sau
cos bình 3x=3/4
tìm min max hàm số
y=4/căn3 sinx+cosx
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Toán Lớp 11: giải pt sau
cos bình 3x=3/4
tìm min max hàm số
y=4/căn3 sinx+cosx
Comments ( 1 )
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}$