Giúp mình với mấy bạn

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Giúp mình với mấy bạn

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9 months 2021-04-20T16:06:16+00:00 1 Answers 19 views 0

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