Дано уравнение   |cosx| = -sqrt3 * sinx

1)сosx<0x∈(π/2+2πn;3π/2+2πn,n∈z)-cosx=-√3sinxcosx-√3sinx=02(1/2cosx-√3/2sinx)=02cos(x+π/3)=0x+π/3=π/2+πn,n∈zx=π/6+πn,n∈zнет решения2)сosx≥0x∈[-π/2+2πn;π/2+2πn,n∈z]cosx=-√3sinxcos(x-π/3)=0x-π/3=π/2+πnx=5π/6+πn,n∈z