Nghiệm phương trình 2{sin}}x{sin}}2x = 3 - căn 3 {sinx} có dạng

Đáp án

\(a + b = 4\)

Giải thích

\(2{\rm{sin}}x{\rm{sin}}2x = 3 - \sqrt 3 {\rm{sinx}}\)

\( \Leftrightarrow {\rm{cos}}x - {\rm{cos}}3x + \sqrt 3 {\rm{sin}}x = 3\)

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

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{1}{2}{\rm{cos}}x + \frac{{\sqrt 3 }}{2}{\rm{sin}}x = 1 \Leftrightarrow {\rm{cos}}\left( {x - \frac{\pi }{3}} \right) = 1}\\{{\rm{cos}}3x = - 1 \Leftrightarrow 3x = \pi + k2\pi }\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\end{array} \Leftrightarrow x = \frac{\pi }{3} + k2\pi } \right.} \right.\)

Vậy \(a = 1,b = 3 \Rightarrow a + b = 4\)