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

а)√2 sin x - √2 cos = √32(√2/2sinx-√2/2сjsx)=√32(sinxcosπ/4-sinπ/4cosx)=√3sin(x-π/4)=√√3/2x-π/4=(-1)^n*π/3+πn,n∈zx=π/4+(-1)^n*π/3+πn,n∈z b)1 + cos x = ctg x/22cos²x/2=(cosx/2)/sinx/2sinx/2≠02cos²x/2sinx/2-cosx/2=0cosx/2*(2cosx/2sinx/2-1)=0cosx/2=0⇒x/2=π/2+πn,n∈z⇒x=π+2πn,n∈zsinx-1=0⇒sinx=1⇒x=π/2+2πk,k∈z