решите уравнение |cosx| = - (sqrt{3}) * sinx [2p; 7p/2]

1) cosx<0⇒x∈ (π/2+2πn; 3π/2+2πn, n∈z)

-cosx+√3sinx=0

2 (√3/2sinx-1/2cosx) = 0

2sin (x-π/6) = 0

x-π/6=πn

x=π/6+πn U x∈ (π/2+2πn; 3π/2+2πn, n∈z) ⇒x=7π/6+2πn

2π≤7π/6+2πn≤7π/2

12≤7+12n≤21

5≤12n≤14

5/12≤n≤7/6

n=1⇒x=7π/6+2π=19π/6

2) cosx≥0⇒x∈[-π/2+2πk; π/2+2πk, k∈z]

cosx+√3sinx=0

2sin (x+π/6) = 0

x+π/6=πk

x=-π/6+πk U x∈[-π/2+2πk; π/2+2πk, k∈z]⇒x=π/6+2πk

2π≤π/6+2πk≤7π/2

12≤1+12k≤21

11≤12k≤20

11/12≤k≤5/3

k=1⇒x=π/6+2π=13π/6