Giải phương trình \[4\sin x\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{{2\pi }}{3}} \right) + \cos 3x = 1\]

Trả lời:

\[4\sin x\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{{2\pi }}{3}} \right) + \cos 3x = 1\]

\[ \Leftrightarrow 4sinx.( - \frac{1}{2})[cos(2x + \pi ) - cos( - \frac{\pi }{3})] + cos3x = 1\]

\[ \Leftrightarrow - 2sinx( - cos2x - \frac{1}{2}) + cos3x = 1\]

\[ \Leftrightarrow 2sinxcos2x + sinx + cos3x = 1\]

\[ \Leftrightarrow sin3x - sinx + sinx + cos3x = 1\]

\[ \Leftrightarrow sin3x + cos3x = 1\]

\[ \Leftrightarrow \frac{1}{{\sqrt 2 }}sin3x + \frac{1}{{\sqrt 2 }}cos3x = \frac{1}{{\sqrt 2 }}\]

\[ \Leftrightarrow sin(3x + \frac{\pi }{4}) = sin\frac{\pi }{4}\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi }\\{3x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{k2\pi }}{3}}\\{x = \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.(k \in \mathbb{Z})\)

Vậy phương trình có nghiệm là:

\(\left[ {\begin{array}{*{20}{c}}{x = \frac{{k2\pi }}{3}}\\{x = \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.(k \in \mathbb{Z})\)

Đáp án cần chọn là: B