2cos^2x*(cos2x-4sinx+1)+3sin^2x=0

Đáp án:

Giải thích các bước giải:
\[\begin{array}{l}
2{\cos ^2}x(\cos 2x – 4\sin x + 1) + 3{\sin ^2}x = 0\\
\Leftrightarrow 2{\cos ^2}x(2{\cos ^2}x – 4\sin x) + 3{\sin ^2}x = 0\\
\Leftrightarrow 4{\cos ^4}x – 8{\cos ^2}x.\sin x + 3{\sin ^2}x = 0\\
\Leftrightarrow 4{\cos ^4}x – 2{\cos ^2}x.\sin x – 6{\cos ^2}x.\sin x + 3{\sin ^2}x = 0\\
\Leftrightarrow 2{\cos ^2}x(2{\cos ^2}x – \sin x) – 3\sin x(2{\cos ^2}x – \sin x) = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x – 3\sin x} \right)(2{\cos ^2}x – \sin x) = 0\\
\Leftrightarrow (2 – 2{\sin ^2}x – 3\sin x)(2 – 2{\sin ^2}x – \sin x) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\sin x = – 2 = > vn\\
\sin x = \frac{{ – 1 – \sqrt {17} }}{4} = > vn\\
\sin x = \frac{{ – 1 + \sqrt {17} }}{4} \Leftrightarrow \left[ \begin{array}{l}
x = \arcsin \frac{{ – 1 + \sqrt {17} }}{4} + k2\pi \\
x = \pi – \arcsin \frac{{ – 1 + \sqrt {17} }}{4} + k2\pi
\end{array} \right.
\end{array} \right.\\

\end{array}\]