2cos²x=√3cosx;
2cos²x-√3cosx=0;
cosx(2cosx-√3)=0;
cosx=0;
x=π/2+πn, n∈Z;
π≤π/2+πn≤3π;
π-π/2≤πn≤3π-π/2;
π/2≤πn≤5π/2;
1/2≤n≤5/2;
n=1;2.
n=1: x=π/2+π=3π/2;
n=2: x=π/2+2π=5π/2.
или
2cosx=√3;
cosx=√3/2;
x=+-π/6+2πk, k∈Z;
π≤π/6+2πk≤3π;
π-π/6≤2πk≤3π-π/6;
5π/6≤2πk≤17π/6;
5/12≤k≤17/12;
k=1.
k=1: x=π/6+2π=13π/6.
π≤-π/6+2k≤3π;
7/12≤k≤19/12;
k=1.
k=1: x=-π/6+2π=11π/6.
Ответ: 3π/2; 11π/6; 13π/6; 5π/2.