$\sin^2x +\sin x \cos 4x +\cos^24x =\frac{3}{4}$
$<=> sin²x + sinx.cos4x + (1/4)cos²4x + (3/4)cos²4x - 3/4 = 0$
$<=> [sinx + (1/2)cos4x]² - (3/4)sin²4x = 0$
$<=> [sinx+(1/2)cos4x -(√3/2)sin4x][sinx+(1/2)cos4x+(√3/2)sin4x] = 0$
$<=> [sinx - sin(4x-pi/6)][sinx + sin(4x+pi/6)] = 0$
$<=>$
$[sin(4x-pi/6) = sinx$
$[sin(4x+pi/6) = sin(-x)$
$<=>$
$[4x-pi/6 = x + 2kpi$
$[4x-pi/6 = pi-x+2kpi$
$[4x+pi/6 = -x + 2kpi$
$[4x+pi/6 = pi+x+2kpi$
$<=>$
$[x = pi/18 + 2kpi/3$
$[x = 7pi/30 + 2kpi/5$
$[x = -pi/30 + 2kpi/5$
$[x = 5pi/18 + 2kpi/3$ ($k$$∈$ $Z$)
