Question

2sin ^ 2 (3п / 2-x) = cosx
[3п / 2; 0]

Answer 1

2сcos²x-cosx=0
cosx*(2cosx-1)=0
cosx=0⇒x=π/2+πn,n∈z
0≤π/2+πn≤3π/2
0≤1+2n≤3
-1≤2n≤2
-1/2≤n≤1
n=0⇒x=π/2
n=1⇒x=π/2+π=3π/2
2cosx-1=0
cosx=1/2
x=-π/3+2πk,k∈z U x=π/3+2πt,t∈z
0≤-π/3+2πk≤3π/2
0≤-2+12k≤9
2≤12k≤11
1/6≤k≤11/12
нет решения
0≤π/3+2πt≤3π/2
0≤2+12t≤9
-2≤12t≤7
-1/6≤t≤7/12
t=0⇒x=π/3