Question

1)4sin$^{2}$2x=3
(4синус в квадрате двух икс равно 3)

2)cos2x-6sinxcosx+3=arccos(-$frac{1}{2}$) - $frac{2}{3} pi$

Answer 1

4sin^2(2x)=3
sin^2x(2x)=3/4
sinx=√3/2  x=(-1)^k*П/3+Пk
sinx=-√3/2  x=(-1)^(k+1)*П/3+Пk
2)cos2x-6sinxcosx+3=arccos(-1/2)-2П/3
cos2x-6sinxcosx=2п/3-2П/3
cos2x-3sin2x=0
tg2x=1/3
x=1/2arctg(1/3)+Пk/2

Answer 2

$4sin^2(2x)=3$
$sin^2(2x)=3/4$

$sinx= sqrt{3} /2 \ x = (-1)^n* frac{ pi }{3}+ pi n, n in Z$

$sinx= -sqrt{3} /2 \ x = (-1)^{n+1}* frac{ pi }{3}+ pi n, n in Z$