Giải phương trình \(\left( {2\sin...

A

- Hướng dẫn giải

Lời giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\left( {2\sin x - 1} \right)\left( {2\cos 2x + 2\sin x + 1} \right) = 3 - 4{\cos ^2}x\\ \Leftrightarrow \left( {2\sin x - 1} \right)\left( {2 - 4{{\sin }^2}x + 2\sin x + 1} \right) = 3 - 4{\cos ^2}x\\ \Leftrightarrow \left( {2\sin x - 1} \right)\left( { - 4{{\sin }^2}x + 2\sin x + 3} \right) = 3 - 4{\cos ^2}x\\ \Leftrightarrow  - 8{\sin ^3}x + 4{\sin ^2}x + 6\sin x + 4{\sin ^2}x - 2\sin x - 3 = 3 - 4{\cos ^2}x\\ \Leftrightarrow  - 8{\sin ^3}x + 8{\sin ^2}x + 4\sin x - 3 = 3 - 4\left( {1 - {{\sin }^2}x} \right)\\ \Leftrightarrow  - 8{\sin ^3}x + 8{\sin ^2}x + 4\sin x - 3 = 3 - 4 + 4{\sin ^2}x\\ \Leftrightarrow  - 8{\sin ^3}x + 4{\sin ^2}x + 4\sin x - 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{{\sqrt 2 }}{2}\\\sin x =  - \frac{{\sqrt 2 }}{2}\\\sin x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{{3\pi }}{4} + k2\pi \\x = \frac{{ - \pi }}{4} + k2\pi \\x = \frac{{5\pi }}{4} + k2\pi \\x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Hợp nghiệm \( \Rightarrow S = \left\{ {\frac{\pi }{4} + \frac{{k\pi }}{2};\frac{\pi }{6} + k2\pi ,\frac{{5\pi }}{6} + k2\pi ,\,\,\kappa  \in \mathbb{Z}} \right\}\).

Chọn A.