[tex]2sinx.sin3x= m - 2m^{2} +1[/tex] có nghiệm [tex]\frac{-\pi }{4} < x< \frac{\pi }{4} [tex][/tex][/tex]
$2 \sin x \cdot \sin 3x= m - 2m^{2} +1$
$\implies \cos 2x - \cos 4x = m - 2m^{2} +1$ (Vì $2 \cdot \sin \theta \cdot \sin \phi = \cos (\theta - \phi) - \cos (\theta + \phi)$)
$\implies \cos 2x - 2\cos^2 2x +1 = m - 2m^{2} +1$ (Vì $\cos 2\theta= 2\cos^2 \theta - 1$)
$\implies (\cos 2x - m)(2 \cos 2x + 2m - 1) = 0$
$\implies \cos 2x = m$ hoặc $\cos 2x = \dfrac{1 - 2m}{2}$
$-\dfrac{\pi}{4} < x < \dfrac{\pi}{4} \implies -\dfrac{\pi}{2} < 2x < \dfrac{\pi}{2} \implies 0 < \cos 2x \leq 1$
Do đó $0 < m \leq 1 \, (1)$ hoặc $0 < \dfrac{1 - 2m}{2} \leq 1 \implies -\dfrac{1}{2} \leq m < \dfrac{1}{2} \, (2)$
Từ $(1)$ và $(2)$ ra: $-\dfrac{1}{2} \leq m \leq 1$