2sin ^ 2 (3 п / 2-x) = cosx

[3 п / 2; 0]

2 сcos²x-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