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Đáp án:
b) \(\left[ \begin{array}{l}
x = k2\pi \\
x = \arccos \dfrac{1}{6} + k2\pi \\
x = – \arccos \dfrac{1}{6} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4{\cos ^2}x – 2\cos x – 2 = 0\\
\to 2\left( {2\cos x + 1} \right)\left( {\cos x – 1} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = – \dfrac{1}{2}\\
\cos x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = – \dfrac{{2\pi }}{3} + k2\pi \\
x = k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b)6{\sin ^2}x + 7\cos x – 7 = 0\\
\to 6\left( {1 – {{\cos }^2}x} \right) + 7\cos x – 7 = 0\\
\to – 6{\cos ^2}x + 7\cos x – 1 = 0\\
\to \left( {1 – \cos x} \right)\left( {6\cos x – 1} \right) = 0\\
\to \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{1}{6}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = k2\pi \\
x = \arccos \dfrac{1}{6} + k2\pi \\
x = – \arccos \dfrac{1}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
c)2{\sin ^2}x – 4\cos x\sin x – 3{\cos ^2}x = – 2\\
\to \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} – \dfrac{{4\sin x}}{{\cos x}} – 3 = – \dfrac{2}{{{{\cos }^2}x}}\\
\to 2{\tan ^2}x – 4\tan x – 3 = – 2\left( {1 + {{\tan }^2}x} \right)\\
\to 4{\tan ^2}x – 4\tan x – 1 = 0\\
\to \left[ \begin{array}{l}
\tan x = \dfrac{{1 + \sqrt 2 }}{2}\\
\tan x = \dfrac{{1 – \sqrt 2 }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \arctan \left( {\dfrac{{1 + \sqrt 2 }}{2}} \right) + k\pi \\
x = \arctan \left( {\dfrac{{1 – \sqrt 2 }}{2}} \right) + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)