Hỏi đáp Lớp 11

Ta biến đổi phương trình đã cho như sau (để ý: \(\sin 4x = 4\sin x\cos x\cos 2x\)):

\(\eqalign{
& \left( {\cos x + \cos 3x} \right) + 2\cos \left( {x + 4x} \right) = 0 \cr&\Leftrightarrow 2\cos 2x\cos x + 2\left( {\cos 4x\cos x - \sin 4x\sin x} \right) = 0 \cr 
& \Leftrightarrow 2\cos 2x\cos x + 2\cos 4x\cos x - 8{\sin ^2}x\cos x\cos 2x = 0 \cr 
& \Leftrightarrow 2\cos x\left( {\cos 2x + \cos 4x - 4{{\sin }^2}x\cos 2x} \right) = 0 \cr 
& \Leftrightarrow 2\cos x\left[ {\cos 2x + \left( {2{{\cos }^2}x - 1} \right) - 2\left( {1 - \cos 2x} \right)\cos 2x} \right] = 0 \cr 
& \Leftrightarrow 2\cos x\left( {4{{\cos }^2}2x - \cos 2x - 1} \right) = 0 \cr} \)

+) \(\cos x = 0 \) \(\Leftrightarrow x = {\pi  \over 2} + k\pi \)

+) \(4{\cos ^2}x - \cos 2x - 1 = 0 \) \(\Leftrightarrow \cos 2x = {{1 \pm \sqrt {17} } \over 8}\)

Do \(\left| {{{1 \pm \sqrt {17} } \over 8}} \right| < 1\) nên có các số \(\alpha \) và \(\beta \) sao cho \(\cos \alpha  = {{1 - \sqrt {17} } \over 8}\) và \(\cos \beta  = {{1 + \sqrt {17} } \over 8}\). Từ đó:

\(\cos 2x = {{1 - \sqrt {17} } \over 8} \)\(\Leftrightarrow 2x =  \pm \alpha  + k2\pi \) \( \Leftrightarrow x =  \pm {\alpha  \over 2} + k\pi \)

\(\cos 2x = {{1 + \sqrt {17} } \over 8} \Leftrightarrow 2x =  \pm \beta  + k2\pi  \Leftrightarrow x =  \pm {\beta  \over 2} + k\pi \)

Kết luận: Phương trình đã cho  các  nghiệm \(x = {\pi  \over 2} + k\pi ,x =  \pm {\alpha  \over 2} + k\pi \) và \(x =  \pm {\beta  \over 2} + k\pi ,\)với \(\cos \alpha  = {{1 - \sqrt {17} } \over 8}\) và \(\cos \beta  = {{1 + \sqrt {17} } \over 8}\).

\({\cos ^2}3x + {\cos ^2}4x + {\cos ^2}5x = {3 \over 2}\)

\(\begin{array}{l}
\Leftrightarrow \frac{{1 + \cos 6x}}{2} + \frac{{1 + \cos 8x}}{2} + \frac{{1 + \cos 10x}}{2} = \frac{3}{2}\\
\Leftrightarrow 1 + \cos 6x + 1 + \cos 8x + 1 + \cos 10x = 3\\
\Leftrightarrow \cos 6x + \cos 8x + \cos 10x = 0\\
\Leftrightarrow 2\cos 8x\cos 2x + \cos 8x = 0\\
\Leftrightarrow \cos 8x\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 8x = 0\\
\cos 2x = - \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
8x = \frac{\pi }{2} + k\pi \\
2x = \pm \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{16}} + \frac{{k\pi }}{8}\\
x = \pm \frac{\pi }{3} + k\pi
\end{array} \right.
\end{array}\)

Vậy \(x = {\pi  \over {16}} + {{k\pi } \over 8},x =  \pm {\pi  \over 3} + k\pi \)     

\({\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = 2\)

\(\begin{array}{l}
\Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2}\\
+ \frac{{1 + \cos 6x}}{2} + \frac{{1 + \cos 8x}}{2} = 2\\
\Leftrightarrow 1 + \cos 2x + 1 + \cos 4x\\
+ 1 + \cos 6x + 1 + \cos 8x = 4\\
\Leftrightarrow \cos 2x + \cos 4x + \cos 6x + \cos 8x = 0\\
\Leftrightarrow 2\cos 3x\cos x + 2\cos 7x\cos x = 0\\
\Leftrightarrow 2\cos x\left( {\cos 3x + \cos 7x} \right) = 0\\
\Leftrightarrow 2\cos x.2\cos 5x\cos 2x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos 2x = 0\\
\cos 5x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k\pi \\
x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x = \frac{\pi }{{10}} + \frac{{k\pi }}{5}
\end{array} \right.
\end{array}\)

Vậy \(x = {\pi  \over 2} + k\pi ,x = {\pi  \over 4} + {{k\pi } \over 2},\) \(x = {\pi  \over {10}} + {{k\pi } \over 5}\)

\(\begin{array}{l}
{\sin ^2}2x + {\sin ^2}4x = {\sin ^2}6x\\
\Leftrightarrow \frac{{1 - \cos 4x}}{2} + {\sin ^2}4x = \frac{{1 - \cos 12x}}{2}\\
\Leftrightarrow 1 - \cos 4x + 2{\sin ^2}4x = 1 - \cos 12x\\
\Leftrightarrow 2{\sin ^2}4x + \cos 12x - \cos 4x = 0\\
\Leftrightarrow 2{\sin ^2}4x - 2\sin 8x\sin 4x = 0\\
\Leftrightarrow 2{\sin ^2}4x - 4{\sin ^2}4x\cos 4x = 0\\
\Leftrightarrow 2{\sin ^2}4x\left( {1 - 2\cos 4x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = 0\\
\cos 4x = \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = k\pi \\
4x = \pm \frac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{4}\\
x = \pm \frac{\pi }{{12}} + \frac{{k\pi }}{2}
\end{array} \right.
\end{array}\)

Vậy \(x = {{k\pi } \over 4},x =  \pm {\pi  \over {12}} + {{k\pi } \over 2}\).

\({\sin ^4}x + {\cos ^4}x = \cos 4x\)

\(\begin{array}{l}
\Leftrightarrow {\left( {\frac{{1 - \cos 2x}}{2}} \right)^2} + {\left( {\frac{{1 + \cos 2x}}{2}} \right)^2}\\
= 2{\cos ^2}2x - 1\\
\Leftrightarrow {\left( {1 - \cos 2x} \right)^2} + {\left( {1 + \cos 2x} \right)^2}\\
= 8{\cos ^2}2x - 4\\
\Leftrightarrow 1 - 2\cos 2x + {\cos ^2}2x\\
+ 1 + 2\cos 2x + {\cos ^2}2x\\
= 8{\cos ^2}2x - 4\\
\Leftrightarrow 6{\cos ^2}2x - 6 = 0\\
\Leftrightarrow {\cos ^2}2x = 1\\
\Leftrightarrow \sin 2x = 0\\
\Leftrightarrow 2x = k\pi \\
\Leftrightarrow x = \frac{{k\pi }}{2}
\end{array}\)

Vậy \(x = {{k\pi } \over 2}\)

\(\begin{array}{l}
\sqrt 3 \sin {15^0} + \cos {15^0} - \sqrt 2 \\
= 2\left( {\frac{{\sqrt 3 }}{2}\sin {{15}^0} + \frac{1}{2}\cos {{15}^0}} \right) - \sqrt 2 \\
= 2\left( {\cos {{30}^0}\sin {{15}^0} + \sin {{30}^0}\cos {{15}^0}} \right) - \sqrt 2 \\
= 2\sin \left( {{{15}^0} + {{30}^0}} \right) - \sqrt 2 \\
= 2\sin {45^0} - \sqrt 2 \\
= 2.\frac{{\sqrt 2 }}{2} - \sqrt 2 \\
= 0
\end{array}\)

\(\begin{array}{l}
\frac{1}{{\sin \frac{\pi }{9}}} - \frac{1}{{\sqrt 3 \cos \frac{\pi }{9}}}\\
= \frac{{\sqrt 3 \cos \frac{\pi }{9} - \sin \frac{\pi }{9}}}{{\sqrt 3 \cos \frac{\pi }{9}\sin \frac{\pi }{9}}}\\
= \frac{{2\left( {\frac{{\sqrt 3 }}{2}\cos \frac{\pi }{9} - \frac{1}{2}\sin \frac{\pi }{9}} \right)}}{{\frac{{\sqrt 3 }}{2}.2\cos \frac{\pi }{9}\sin \frac{\pi }{9}}}\\
= \frac{{2\left( {\sin \frac{\pi }{3}\cos \frac{\pi }{9} - \cos \frac{\pi }{3}\sin \frac{\pi }{9}} \right)}}{{\frac{{\sqrt 3 }}{2}.\sin \frac{{2\pi }}{9}}}\\
= \frac{{2\sin \frac{{2\pi }}{9}}}{{\frac{{\sqrt 3 }}{2}.\sin \frac{{2\pi }}{9}}}\\
= \frac{4}{{\sqrt 3 }}
\end{array}\)

Ta có:

\(\begin{array}{l}
{\left( {2 - \sqrt 3 } \right)^2} + {1^2} = 4 - 4\sqrt 3 + 3 + 1\\
= 8 - 4\sqrt 3 = 4\left( {2 - \sqrt 3 } \right)\\
\Rightarrow y = 2\sqrt {2 - \sqrt 3 } \left[ {\frac{{\sqrt {2 - \sqrt 3 } }}{2}\sin 2x + \frac{1}{{2\sqrt {2 - \sqrt 3 } }}\cos 2x} \right]\\
= 2\sqrt {2 - \sqrt 3 } \sin \left( {2x + \alpha } \right)
\end{array}\)

với \(\alpha \) thỏa mãn 

\(\left\{ \begin{array}{l}
\cos \alpha = \frac{{\sqrt {2 - \sqrt 3 } }}{2}\\
\sin \alpha = \frac{1}{{2\sqrt {2 - \sqrt 3 } }}
\end{array} \right.\)

Do đó \( - 2\sqrt {2 - \sqrt 3 }  \le y \le 2\sqrt {2 - \sqrt 3 } \)

Vậy giá trị lớn nhất là \(2\sqrt {2 - \sqrt 3 } ,\) giá trị nhỏ nhất là  \( - 2\sqrt {2 - \sqrt 3 } .\)

Ta có:

\(y = {\left( {\sin x - \cos x} \right)^2} + 2\cos 2x + 3\sin x\cos x \)

\(\begin{array}{l}
= {\sin ^2}x - 2\sin x\cos x + {\cos ^2}x + 2\cos 2x + \frac{3}{2}.2\sin x\cos x\\
= 1 - \sin 2x + 2\cos 2x + \frac{3}{2}\sin 2x
\end{array}\)

\(= 1 + {1 \over 2}\sin 2x + 2\cos 2x.\)

\(\begin{array}{l}
\frac{1}{2}\sin 2x + 2\cos 2x\\
= \frac{{\sqrt {17} }}{2}\left( {\frac{1}{{\sqrt {17} }}\sin 2x + \frac{4}{{\sqrt {17} }}\cos 2x} \right)\\
= \frac{{\sqrt {17} }}{2}\sin \left( {2x + \alpha } \right)
\end{array}\)

với \(\alpha\) thỏa mãn 

\({\left\{ \begin{array}{l}
\cos \alpha = \frac{1}{{\sqrt {17} }}\\
\sin \alpha = \frac{4}{{\sqrt {17} }}
\end{array} \right.}\)

Mà \( - 1 \le \sin \left( {2x + \alpha } \right) \le 1\) nên \( - \frac{{\sqrt {17} }}{2} \le \frac{{\sqrt {17} }}{2}\sin \left( {2x + \alpha } \right) \le \frac{{\sqrt {17} }}{2}\)

Do đó:

\(\begin{array}{l}
- \frac{{\sqrt {17} }}{2} \le \frac{1}{2}\sin 2x + 2\cos 2x \le \frac{{\sqrt {17} }}{2}\\
\Rightarrow 1 - \frac{{\sqrt {17} }}{2} \le 1 + \frac{1}{2}\sin 2x + 2\cos 2x \le 1 + \frac{{\sqrt {17} }}{2}\\
\Rightarrow 1 - \frac{{\sqrt {17} }}{2} \le y \le 1 + \frac{{\sqrt {17} }}{2}
\end{array}\)

Vậy giá trị lớn nhất và giá trị nhỏ nhất của y theo thứ tự là \(1 + {{\sqrt {17} } \over 2}\) và  \(1 - {{\sqrt {17} } \over 2}\).