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