|cosx| / cosx - 2 = 2sinx
1)cosx>0 ⇔ x∈(-π/2+2πn; π/2+2πn), n∈Z
|cosx| / cosx - 2 = 2sinx ⇒ cosx / cosx - 2 = 2sinx ⇒1-2=2sinx
!!!!! x∈(-π/2+2πn; π/2+2πn) !!!!!
⇒sinx =-1/2 ⇒ x1=-π/6+2πk, x2=-π+π/6+2πk; n,k∈Z
x1∈(-π/2+2πn; π/2+2πn) x2∉(-π/2+2πn; π/2+2πn)
1)x1=-π/6+2πk, k∈Z
2)cosx<0 ⇔ x∈(π/2+2πn; 3π/2<span>+2πn)
|cosx| / cosx - 2 = 2sinx ⇒ -cosx / cosx - 2 = 2sinx ⇒-1-2=2sinx
⇒sinx =-3/2 ⇒ нет решений, т.к. |sinx| ≤1, а |-3/2| >1
ответ: x1=-π/6+2πk, k∈Z