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

Giải đáp:

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