Решите уравнение а) 2cos^2x+1=2sqrt2cos(3pi-x) найдите корни б) [3pi/2; 3pi]

А) 2cos²x+1=2√2cos(3π-x)=2√2cosx,
2cos²x-2√2cosx+1=(√2cosx-1)²=0,
√2cosx-1=0,√2cosx=1,cosx=1/√2
x=+-(π/4)+2πn,n∈Z 
 б).найти х∈[3π/2; 3π]-?
n=1, x=π/4+2π=2,25π∈ [3π/2; 3π],х =2π-π/4=1.75π∈ [3π/2; 3π] 
n=2,x=4π-π/43,35π∉ [3π/2; 3π]
ответ:2,25π и  1.75π