Question

PLEASE HELP!! THANK YOU!

Answer 1

Recall that 2sinxcosx = sin2x.
Thus, sinxcosx = sin2x/2

So, $\frac{sin2x}{2} = \frac{1}{4}$
$sin2x = \frac{1}{2}$
$2x = \frac{\pi}{6}, \frac{5\pi}{6}$
$x = \frac{\pi}{12}, \frac{5\pi}{12}$

In general form:
$x = \pi n + \frac{\pi}{12}$
$x = \pi n + \frac{5\pi}{12}$

Answer 2

Answers A and C and correct for this equation.