Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{1}{2}\arccos \dfrac{1}{3} + k\pi \\
x = - \dfrac{1}{2}\arccos \dfrac{1}{3} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\cos x.\cos y = \cos \left( {x + y} \right) + \cos \left( {x - y} \right)\\
\cos 2x = 2{\cos ^2}x - 1\\
\sin \left( {4x - \dfrac{{7\pi }}{2}} \right) - 8\cos x.\cos 3x = 1\\
\Leftrightarrow \sin \left[ {\left( {4x + \dfrac{\pi }{2}} \right) - 4\pi } \right] - 4.\left( {2\cos 3x.\cos x} \right) = 1\\
\Leftrightarrow \sin \left( {4x + \dfrac{\pi }{2}} \right) - 4.\left[ {\cos \left( {3x + x} \right) + \cos \left( {3x - x} \right)} \right] = 1\\
\Leftrightarrow \cos \left[ {\dfrac{\pi }{2} - \left( {4x + \dfrac{\pi }{2}} \right)} \right] - 4.\left( {\cos 4x + \cos 2x} \right) = 1\\
\Leftrightarrow \cos \left( { - 4x} \right) - 4\cos 4x - 4\cos 2x = 1\\
\Leftrightarrow \cos 4x - 4\cos 4x - 4\cos 2x - 1 = 0\\
\Leftrightarrow 3\cos 4x + 4\cos 2x + 1 = 0\\
\Leftrightarrow 3.\left( {2{{\cos }^2}2x - 1} \right) + 4\cos 2x + 1 = 0\\
\Leftrightarrow 6{\cos ^2}2x + 4\cos 2x - 2 = 0\\
\Leftrightarrow 3{\cos ^2}2x + 2\cos 2x - 1 = 0\\
\Leftrightarrow \left( {\cos 2x + 1} \right)\left( {3\cos 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x + 1 = 0\\
3\cos 2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = - 1\\
\cos 2x = \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pi + k2\pi \\
2x = \arccos \dfrac{1}{3} + k2\pi \\
2x = - \arccos \dfrac{1}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{1}{2}\arccos \dfrac{1}{3} + k\pi \\
x = - \dfrac{1}{2}\arccos \dfrac{1}{3} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
