Giải phương trình sau
3sin2x+4cosx−2 = 0
Ta có: 3sin2x+4cosx−2 = 0
\(\begin{array}{l}
\Leftrightarrow 3(1 - {\cos ^2}x) + 4\cos x - 2 = 0\\
\Leftrightarrow 3{\cos ^2}x - 4\cos x - 1 = 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\cos x = \frac{{2 - \sqrt 7 }}{3}}\\
{\cos x = \frac{{2 + \sqrt 7 }}{3} > 1\,\,{\rm{(l)}}}
\end{array}} \right.\\
\Leftrightarrow x = \pm \arccos \left( {\frac{{2 - \sqrt 7 }}{3}} \right) + k2\pi ,k \in Z
\end{array}\)
-- Mod Toán 11