a)$ 4cos^2x(1+sinx)+2\sqrt{3} cosxcos2x = 1+2sinx$
$<=>2cosx(2cosx+sin2x+\sqrt{3}cos2x)=2(sinx+sin\pi/6)$
$<=>2cosx[2cosx+2cos(2x-\pi/6)]=4sin\dfrac{x+\pi/6}{2}cos\dfrac{x-\pi/6}{2}$
$<=>8cosxcos\dfrac{3x-\pi/6}{2}cos\dfrac{x-\pi/6}{2}=4sin\dfrac{x+\pi/6}{2}cos\dfrac{x-\pi/6}{2}$
$<=>\dfrac{x-\pi/6}{2}=\pi/2+k\pi$
$cos(5x/2-\pi/12)+cos(x/2-\pi/12)=sin(x/2+\pi/12)=-cos(x/2+7\pi/12)$
$<=>cos(5x/2-\pi/12)-cos(x+\pi/12)=0$