$ 2\cos 4x = 2\cos^2 3x+ 2m\sin^2 x$. Trong đó+ $\cos 4x= 2\cos^2 2x-1$+ $ 2\cos^2 3x=1+\cos6 x=1+4\cos^3 2x-3\cos 2x$+ $2\sin^2 x=1-\cos 2x$. Thay vào ta có$\Leftrightarrow 4\cos^3 2x-4\cos^2 x - (m+3)\cos 2x +m+3=0$$\Leftrightarrow (\cos 2x-1)(4\cos^2 x- m-3)=0$+ $\cos 2x=1$ (không thuộc $ (0;\ \dfrac{\pi}{12})$+ $4\cos^2 x=m+3$$\Leftrightarrow \cos 4x = \dfrac{m+1}{2}$Vì $x \in (0;\ \dfrac{\pi}{12})$ nên $4x \in (0;\ \dfrac{\pi}{3}) \Rightarrow cos4x \in (\dfrac{1}{2};1)$$\Rightarrow \dfrac{1}{2}<\dfrac{m+1}{2}<1 \Leftrightarrow 0
$ 2\cos 4x = 2\cos^2 3x+ 2m\sin^2 x$. Trong đó$\Leftrightarrow \cos 4x= 2\cos^2 2x-1$$\Leftrightarrow 2\cos^2 3x=1+\cos6 x=1+4\cos^3 2x-3\cos 2x$$\Leftrightarrow 2\sin^2 x=1-\cos 2x$. Thay vào ta có$\Leftrightarrow 4\cos^3 2x-4\cos^2 x - (m+3)\cos 2x +m+3=0$$\Leftrightarrow (\cos 2x-1)(4\cos^2 x- m-3)=0$+ $\cos 2x=1$ (không thuộc $ (0;\ \dfrac{\pi}{12})$+ $4\cos^2 x=m+3$$\Leftrightarrow \cos 4x = \dfrac{m+1}{2}$Vì $x \in (0;\ \dfrac{\pi}{12})$ nên $4x \in (0;\ \dfrac{\pi}{3}) \Rightarrow cos4x \in (\dfrac{1}{2};1)$$\Rightarrow \dfrac{1}{2}<\dfrac{m+1}{2}<1 \Leftrightarrow 0<m<1$
$ 2\cos 4x = 2\cos^2 3x+ 2m\sin^2 x$. Trong đó
+ $\cos 4x= 2\cos^2 2x-1$
+ $ 2\cos^2 3x=1+\cos6 x=1+4\cos^3 2x-3\cos 2x$
+ $2\sin^2 x=1-\cos 2x$. Thay vào ta có$\Leftrightarrow 4\cos^3 2x-4\cos^2 x - (m+3)\cos 2x +m+3=0$$\Leftrightarrow (\cos 2x-1)(4\cos^2 x- m-3)=0$+ $\cos 2x=1$ (không thuộc $ (0;\ \dfrac{\pi}{12})$+ $4\cos^2 x=m+3$$\Leftrightarrow \cos 4x = \dfrac{m+1}{2}$Vì $x \in (0;\ \dfrac{\pi}{12})$ nên $4x \in (0;\ \dfrac{\pi}{3}) \Rightarrow cos4x \in (\dfrac{1}{2};1)$$\Rightarrow \dfrac{1}{2}<\dfrac{m+1}{2}<1 \Leftrightarrow 0