Đáp án:
\[x = \dfrac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\sin x.\sin y = - \left[ {\cos \left( {x + y} \right) - \cos \left( {x - y} \right)} \right]\\
\cos 2x = 2{\cos ^2}x - 1\\
2\sin x.\sin 3x + \cos 2x = - 3\\
\Leftrightarrow - \left[ {\cos \left( {x + 3x} \right) - \cos \left( {x - 3x} \right)} \right] + \cos 2x = - 3\\
\Leftrightarrow - \left( {\cos 4x - \cos \left( { - 2x} \right)} \right) + \cos 2x = - 3\\
\Leftrightarrow - \left( {\cos 4x - \cos 2x} \right) + \cos 2x = - 3\\
\Leftrightarrow - \cos 4x + \cos 2x + \cos 2x = - 3\\
\Leftrightarrow - \left( {2{{\cos }^2}2x - 1} \right) + 2\cos 2x + 3 = 0\\
\Leftrightarrow - 2{\cos ^2}2x + 1 + 2\cos 2x + 3 = 0\\
\Leftrightarrow - 2{\cos ^2}2x + 2\cos 2x + 4 = 0\\
\Leftrightarrow {\cos ^2}2x - \cos 2x - 2 = 0\\
\Leftrightarrow \left( {\cos 2x - 2} \right)\left( {\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x - 2 = 0\\
\cos 2x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 2\\
\cos 2x = - 1
\end{array} \right.\\
- 1 \le \cos 2x \le 1 \Rightarrow \cos 2x = - 1\\
\cos 2x = - 1\\
\Leftrightarrow 2x = \pi + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)
\end{array}\)
