Đáp án:
\(\left[ \begin{array}{l}
x = \pm \arccos \dfrac{{\sqrt 5 }}{5} + k2\pi \\
x = \pm \arccos \dfrac{{ - \sqrt 5 }}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + 2\sin x = 4\cos x + \sin 2x\\
\Leftrightarrow {\sin ^2}x + 2\sin x = 4\cos x + 2\sin x.\cos x\\
\Leftrightarrow \sin x.\left( {\sin x + 2} \right) = 2\cos x.\left( {2 + \sin x} \right)\\
\Leftrightarrow \sin x.\left( {\sin x + 2} \right) - 2\cos x.\left( {\sin x + 2} \right) = 0\\
\Leftrightarrow \left( {\sin x + 2} \right)\left( {\sin x - 2\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + 2 = 0\\
\sin x - 2\cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = - 2\\
\sin x = 2\cos x
\end{array} \right.\\
- 1 \le \sin x \le 1 \Rightarrow \sin x = 2\cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\left( {2\cos x} \right)^2} + {\cos ^2}x = 1\\
\Leftrightarrow 4{\cos ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow 5{\cos ^2}x = 1\\
\Leftrightarrow {\cos ^2}x = \dfrac{1}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 5 }}{5}\\
\cos x = - \dfrac{{\sqrt 5 }}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pm \arccos \dfrac{{\sqrt 5 }}{5} + k2\pi \\
x = \pm \arccos \dfrac{{ - \sqrt 5 }}{5} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
