Sin2x = 1 + √2cosx + cos2x

Đáp án:

$\begin{array}{l}
\sin 2x = 1 + \sqrt 2 .\cos x + \cos 2x\\
 \Rightarrow 2.\sin x.\cos x = 1 + \sqrt 2 .\cos x + 2{\cos ^2}x - 1\\
 \Rightarrow 2\sin x.\cos x = \sqrt 2 .cosx + 2co{s^2}x\\
 \Rightarrow \cos x.\left( {2\sin x - \sqrt 2  - 2\cos x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cos x = 0\\
2.\left( {\sin x - \cos x} \right) = \sqrt 2 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x - \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k2\pi \\
x = \dfrac{{13\pi }}{{12}} + k2\pi 
\end{array} \right.
\end{array}$