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:
\(\begin{array}{l}
1)\sin x.\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = - \dfrac{1}{8}\\
\to \dfrac{{\sin 2x}}{2}.\cos 2x = - \dfrac{1}{8}\\
\to \dfrac{{\sin 2x.\cos 2x}}{2} = - \dfrac{1}{8}\\
\to \dfrac{{\sin 4x}}{4} = - \dfrac{1}{8}\\
\to \sin 4x = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
4x = - \dfrac{\pi }{6} + k2\pi \\
4x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
2)\cos 2x + \cos x + 1 = 0\\
\to 2{\cos ^2}x - 1 + \cos x + 1 = 0\\
\to 2{\cos ^2}x + \cos x = 0\\
\to \cos x\left( {2\cos x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = 0\\
\cos x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
3)6{\cos ^2}x - \cos x - 1 = 0\\
\to \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x = - \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi \\
x = \arccos \left( { - \dfrac{1}{3}} \right) + k2\pi \\
x = - \arccos \left( { - \dfrac{1}{3}} \right) + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
4)6{\cos ^2}x + 5\sin x - 7 = 0\\
\to 6 - 6{\sin ^2}x + 5\sin x - 7 = 0\\
\to - 6{\sin ^2}x + 5\sin x - 1 = 0\\
\to \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \arcsin \dfrac{1}{3} + k2\pi \\
x = \pi - \arcsin \dfrac{1}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
