Nghiệm của phương trình lượng giác: sin x − √ 3 cos x = 2 cos 2x là (với k ∈ Z )

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Nghiệm của phương trình lượng giác: \({\rm{sin}}x - \sqrt 3 {\rm{cos}}x = 2{\rm{cos}}2x\) là (với \(k \in \mathbb{Z}\))

\(\left[ {\begin{array}{*{20}{l}}{x = \frac{\pi }{6} + k2\pi }\\{x = \frac{{ - \pi }}{{18}} + \frac{{k2\pi }}{3}}\end{array}} \right.\).

\(\left[ {\begin{array}{*{20}{l}}{x = \frac{{5\pi }}{6} + k2\pi }\\{x = \frac{{ - 5\pi }}{{18}} + \frac{{k2\pi }}{3}}\end{array}} \right.\).

\(\left[ {\begin{array}{*{20}{l}}{x = \frac{{ - 5\pi }}{6} + k2\pi }\\{x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}}\end{array}} \right.\).

\(\left[ {\begin{array}{*{20}{l}}{x = \frac{{5\pi }}{6} + k2\pi }\\{x = \frac{{ - 5\pi }}{{18}} + \frac{{k2\pi }}{3}}\end{array}} \right.\).

Giải thích

Ta có: \({\rm{sin}}x - \sqrt 3 {\rm{cos}}x = 2{\rm{cos}}2x\)

\( \Leftrightarrow 2\left( {\frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}{\rm{cos}}x} \right) = 2{\rm{cos}}2x\)

\( \Leftrightarrow \sin \frac{\pi }{6}\sin x - \cos \frac{\pi }{6}.{\rm{cos}}x = {\rm{cos}}2x\)

\( \Leftrightarrow - {\rm{cos}}\left( {x + \frac{\pi }{6}} \right) = {\rm{cos}}2x\)

\( \Leftrightarrow {\rm{cos}}\left( {\pi - x - \frac{\pi }{6}} \right) = {\rm{cos}}2x\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\pi - x - \frac{\pi }{6} = 2x + k2\pi }\\{\pi - x - \frac{\pi }{6} = - 2x + k2\pi }\end{array},\left( {k \in \mathbb{Z}} \right)} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{5\pi }}{{18}} + k\frac{{2\pi }}{3}}\\{x = \frac{{ - 5\pi }}{6} + k2\pi }\end{array},\left( {k \in \mathbb{Z}} \right)} \right.\). Chọn C.