Найти все корни уравнения 2cos^2 x - cos x=0 на промежутке [ -П/2; П]

прилагаю решение в виде фото ;

2cos²x-cos x=0 на промежутке [ -π/2;π]

cosx*(2cos x-1)=0

cosx=0⇒х=π/2+πn; n∈Z

2cos x-1=0

cos x=1/2 ; x=±arccos0.5+2πn; n∈Z

x=±π/3+2πn; n∈Z

1)-π/2≤π/2+πn≤π;

-π≤πn≤π/2

-1≤n≤n≤1/2⇒n=-1; х=π/2-π;   х=-π/2

n=0; х=π/2;   х=π/2

2) -π/2≤-π/3+2πn≤π

π/3-π/2≤2πn≤π+π/3

-π/6≤2πn≤4π/3

-1/12≤n≤2/3

n=0⇒х=-π/3

3)-π/2≤π/3+2πn≤π

-π/3-π/2≤2πn≤π-π/3

-5π/6≤2πn≤2π/3

-5/12≤n≤1/3

n=0⇒х=π/3

ответ х=±π/3; х=±π/2