Решить букву а)..........

A)
$\frac{1}{2}cos2x+ \frac{ \sqrt{3} }{2}sin2x =1 \\ \\ \frac{1}{2}(cos^2x-sin^2x)+ \frac{ \sqrt{3} }{2} *2sinxcosx=sin^2x+cos^2x \\ \\ 0.5cos^2x-cos^2x-0.5sin^2x-sin^2x+ \sqrt{3}sinxcosx=0 \\ -0.5cos^2x-1.5sin^2x+ \sqrt{3} sinxcosx=0 \\ \frac{-0.5cos^2x}{cos^2x}- \frac{1.5sin^2x}{cos^2x}+ \frac{ \sqrt{3}sinxcosx }{cos^2x}= \frac{0}{cos^2x} \\ \\ -0.5-1.5tg^2x+ \sqrt{3}tgx=0 \\ 1.5tg^2x- \sqrt{3}tgx+0.5=0 \\ \\$

$y=tgx \\ 1.5y^2- \sqrt{3}y+0.5=0 \\ \\ D=3-4*1.5*0.5=0 \\ \\ y= \frac{ \sqrt{3} }{3}$

$tgx= \frac{ \sqrt{3} }{3} \\ \\ x= \frac{ \pi }{6}+ \pi k$ ,
k∈Z.

Ответ: $\frac{ \pi }{6}+ \pi k,$ k∈Z.