Respuesta :
Recall that 2sinxcosx = sin2x.
Thus, sinxcosx = sin2x/2
So, [tex]\frac{sin2x}{2} = \frac{1}{4}[/tex]
[tex]sin2x = \frac{1}{2}[/tex]
[tex]2x = \frac{\pi}{6}, \frac{5\pi}{6}[/tex]
[tex]x = \frac{\pi}{12}, \frac{5\pi}{12}[/tex]
In general form:
[tex]x = \pi n + \frac{\pi}{12}[/tex]
[tex]x = \pi n + \frac{5\pi}{12}[/tex]
Thus, sinxcosx = sin2x/2
So, [tex]\frac{sin2x}{2} = \frac{1}{4}[/tex]
[tex]sin2x = \frac{1}{2}[/tex]
[tex]2x = \frac{\pi}{6}, \frac{5\pi}{6}[/tex]
[tex]x = \frac{\pi}{12}, \frac{5\pi}{12}[/tex]
In general form:
[tex]x = \pi n + \frac{\pi}{12}[/tex]
[tex]x = \pi n + \frac{5\pi}{12}[/tex]