Với giá trị nào của n thì đẳng thức sau luôn đúng?

\(\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 12x} } }  = \cos \dfrac{x}{{2n}}\)\(\,\,0 < x < \dfrac{\pi }{{12}}\).

Ta có: \(0 < x < \dfrac{\pi }{{12}} \Rightarrow 0 < \dfrac{{3x}}{2} < 3x < 6x < \dfrac{\pi }{2} \Rightarrow 0 < \cos 6x < \cos 3x < \cos \dfrac{{3x}}{2} < 1\) (do hàm số \(y = \cos x\) là hàm số nghịch biến).

\(\begin{array}{l}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 12x} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}6x - 1} \right)} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + {{\cos }^2}6x - \dfrac{1}{2}} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {{{\cos }^2}6x} } } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 6x} } \left( {do\cos 6x > 0} \right)\\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}3x - 1} \right)} } \\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {{{\cos }^2}3x} } \end{array}\)

\(\begin{array}{l} = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos 3x} \left( {do\cos 3x > 0} \right)\\ = \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\left( {2{{\cos }^2}\dfrac{{3x}}{2} - 1} \right)} \\ = \sqrt {{{\cos }^2}\dfrac{{3x}}{2}}  = \cos \dfrac{{3x}}{2}\left( {do\cos \dfrac{{3x}}{2} > 0} \right)\\ \Rightarrow \cos \dfrac{{3x}}{2} = \cos \dfrac{x}{{2n}}\left( 1 \right)\end{array}\)

 Để (1) luôn đúng \( \Rightarrow \dfrac{{3x}}{2} = \dfrac{x}{{2n}} \Leftrightarrow n = \dfrac{1}{3}\)