Đáp án:
\(\left[ \begin{array}{l}
x = k\pi \\
x = \arctan 2 + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + \sin 2x + 2{\cos ^2}x = 2\\
\Leftrightarrow {\sin ^2}x + 2.\sin x.\cos x + 2{\cos ^2}x = 2.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
\Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + 2{\cos ^2}x = 2{\sin ^2}x + 2{\cos ^2}x\\
\Leftrightarrow - {\sin ^2}x + 2\sin x.\cos x = 0\\
\Leftrightarrow - \sin x.\left( {\sin x - 2\cos x} \right) = 0\,\,\,\,\left( 1 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x - 2\cos x = 0
\end{array} \right.\,\,\,\,\,\left( * \right)\\
TH1:\,\,\,\cos x = 0\\
\left( 1 \right) \Rightarrow \sin x.\sin x = 0 \Leftrightarrow \sin x = 0\\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos x \ne 0\\
\left( * \right) \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\dfrac{{\sin x - 2\cos x}}{{\cos x}} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\dfrac{{\sin x}}{{\cos x}} - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\tan x = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \arctan 2 + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
