sin4x – cos4x = √3. cos8x

Đáp án:

$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$

Giải thích các bước giải:

$\begin{array}{l}
\sin 4x – \cos 4x = \sqrt 3 \cos 8x\\
 \Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \cos 8x = 0\\
 \Leftrightarrow \sin 4x – \cos 4x – \sqrt 3 \left( {{{\cos }^2}4x – {{\sin }^2}4x} \right) = 0\\
 \Leftrightarrow \sin 4x – \cos 4x + \sqrt 3 \left( {\sin 4x – \cos 4x} \right)\left( {\sin 4x + \cos 4x} \right) = 0\\
 \Leftrightarrow \left( {\sin 4x – \cos 4x} \right)\left( {1 + \sqrt 3 \left( {\sin 4x + \cos 4x} \right)} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 4x – \cos 4x = 0\\
\sin 4x + \cos 4x = \dfrac{{ – 1}}{{\sqrt 3 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{{\sqrt 2 }}\sin 4x – \dfrac{1}{{\sqrt 2 }}\cos 4x = 0\\
\dfrac{1}{{\sqrt 2 }}\sin 4x + \dfrac{1}{{\sqrt 2 }}\cos 4x = \dfrac{{ – 1}}{{\sqrt 6 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {4x + \dfrac{\pi }{4}} \right) = 0\\
\cos \left( {4x – \dfrac{\pi }{4}} \right) = \dfrac{{ – 1}}{{\sqrt 6 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
4x – \dfrac{\pi }{4} = \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi \\
4x – \dfrac{\pi }{4} =  – \arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có các họ nghiệm là:

$x = \dfrac{\pi }{{16}} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} + \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2};x = \dfrac{\pi }{{16}} – \dfrac{1}{4}\arccos \left( {\dfrac{{ – 1}}{{\sqrt 6 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)$