решите уравнение |cosx|= - корень из 3*(sinx) помогите срочно

1)cosx<0⇒x∈(π/2+2πn;3π/2+2πn,n∈z)-cosx+√3sinx=02(√3/2sinx-1/2cosx)=02sin(x-π/6)=0x-π/6=πnx=π/6+πn U x∈(π/2+2πn;3π/2+2πn,n∈z)⇒x=7π/6+2πn2π≤7π/6+2πn≤7π/212≤7+12n≤215≤12n≤145/12≤n≤7/6n=1⇒x=7π/6+2π=19π/62)cosx≥0⇒x∈[-π/2+2πk;π/2+2πk,k∈z]cosx+√3sinx=02sin(x+π/6)=0x+π/6=πkx=-π/6+πk U x∈[-π/2+2πk;π/2+2πk,k∈z]⇒x=π/6+2πk2π≤π/6+2πk≤7π/212≤1+12k≤2111≤12k≤2011/12≤k≤5/3k=1⇒x=π/6+2π=13π/6