Đáp án: 4) \(\left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x = \arcsin \dfrac{1}{3} + k2\pi \\x = \pi – \arcsin \dfrac{1}{3} + k2\pi \end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}1)\sin x.\cos x\left( {{{\cos }^2}x – {{\sin }^2}x} \right) = – \dfrac{1}{8}\\ \to \dfrac{{\sin 2x}}{2}.\cos 2x = – \dfrac{1}{8}\\ \to \dfrac{{\sin 2x.\cos 2x}}{2} = – \dfrac{1}{8}\\ \to \dfrac{{\sin 4x}}{4} = – \dfrac{1}{8}\\ \to \sin 4x = – \dfrac{1}{2}\\ \to \left[ \begin{array}{l}4x = – \dfrac{\pi }{6} + k2\pi \\4x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}\end{array} \right.\left( {k \in Z} \right)\\2)\cos 2x + \cos x + 1 = 0\\ \to 2{\cos ^2}x – 1 + \cos x + 1 = 0\\ \to 2{\cos ^2}x + \cos x = 0\\ \to \cos x\left( {2\cos x + 1} \right) = 0\\ \to \left[ \begin{array}{l}\cos x = 0\\\cos x = – \dfrac{1}{2}\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x = \dfrac{{2\pi }}{3} + k2\pi \\x = – \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\\3)6{\cos ^2}x – \cos x – 1 = 0\\ \to \left[ \begin{array}{l}\cos x = \dfrac{1}{2}\\\cos x = – \dfrac{1}{3}\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = – \dfrac{\pi }{3} + k2\pi \\x = \arccos \left( { – \dfrac{1}{3}} \right) + k2\pi \\x = – \arccos \left( { – \dfrac{1}{3}} \right) + k2\pi \end{array} \right.\left( {k \in Z} \right)\\4)6{\cos ^2}x + 5\sin x – 7 = 0\\ \to 6 – 6{\sin ^2}x + 5\sin x – 7 = 0\\ \to – 6{\sin ^2}x + 5\sin x – 1 = 0\\ \to \left[ \begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x = \dfrac{1}{3}\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x = \arcsin \dfrac{1}{3} + k2\pi \\x = \pi – \arcsin \dfrac{1}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\end{array}\) Reply
Đáp án:
4) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \arcsin \dfrac{1}{3} + k2\pi \\
x = \pi – \arcsin \dfrac{1}{3} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\sin x.\cos x\left( {{{\cos }^2}x – {{\sin }^2}x} \right) = – \dfrac{1}{8}\\
\to \dfrac{{\sin 2x}}{2}.\cos 2x = – \dfrac{1}{8}\\
\to \dfrac{{\sin 2x.\cos 2x}}{2} = – \dfrac{1}{8}\\
\to \dfrac{{\sin 4x}}{4} = – \dfrac{1}{8}\\
\to \sin 4x = – \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
4x = – \dfrac{\pi }{6} + k2\pi \\
4x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
2)\cos 2x + \cos x + 1 = 0\\
\to 2{\cos ^2}x – 1 + \cos x + 1 = 0\\
\to 2{\cos ^2}x + \cos x = 0\\
\to \cos x\left( {2\cos x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = 0\\
\cos x = – \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = – \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
3)6{\cos ^2}x – \cos x – 1 = 0\\
\to \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x = – \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = – \dfrac{\pi }{3} + k2\pi \\
x = \arccos \left( { – \dfrac{1}{3}} \right) + k2\pi \\
x = – \arccos \left( { – \dfrac{1}{3}} \right) + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
4)6{\cos ^2}x + 5\sin x – 7 = 0\\
\to 6 – 6{\sin ^2}x + 5\sin x – 7 = 0\\
\to – 6{\sin ^2}x + 5\sin x – 1 = 0\\
\to \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \arcsin \dfrac{1}{3} + k2\pi \\
x = \pi – \arcsin \dfrac{1}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)