Решите уравнение а)√2 sin x - √2 cos = √3 б) 1 + cos x = ctg x/2

а)√2 sin x - √2 cos = √3 2(√2/2sinx-√2/2сjsx)=√3 2(sinxcosπ/4-sinπ/4cosx)=√3 sin(x-π/4)=√√3/2 x-π/4=(-1)^n*π/3+πn,n∈z x=π/4+(-1)^n*π/3+πn,n∈z  b)1 + cos x = ctg x/2 2cos²x/2=(cosx/2)/sinx/2 sinx/2≠0 2cos²x/2sinx/2-cosx/2=0 cosx/2*(2cosx/2sinx/2-1)=0 cosx/2=0⇒x/2=π/2+πn,n∈z⇒x=π+2πn,n∈z sinx-1=0⇒sinx=1⇒x=π/2+2πk,k∈z