Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{B - a}}{2} + k\pi \\
x = \dfrac{{\pi - B - a}}{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{1}{2}\sin 2x + 2{\cos ^2}x + 1 = 0\\
\to \sin 2x + 4{\cos ^2}x + 2 = 0\\
\to \sin 2x + 2\left( {2{{\cos }^2}x - 1} \right) + 4 = 0\\
\to \sin 2x + 2.\cos 2x + 4 = 0\\
\to \dfrac{1}{{\sqrt 5 }}.\sin 2x + \dfrac{2}{{\sqrt 5 }}.\cos 2x = - \dfrac{4}{{\sqrt 5 }}\\
Đặt:\left\{ \begin{array}{l}
\dfrac{1}{{\sqrt 5 }} = \cos a\\
\dfrac{2}{{\sqrt 5 }} = \sin a\\
- \dfrac{4}{{\sqrt 5 }} = \sin B
\end{array} \right.\\
Pt \to \sin 2x.\cos a + \cos 2x.\sin x = \sin B\\
\to \sin \left( {2x + a} \right) = \sin B\\
\to \left[ \begin{array}{l}
2x + a = B + k2\pi \\
2x + a = \pi - B + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{B - a}}{2} + k\pi \\
x = \dfrac{{\pi - B - a}}{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
