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