Question

Решите уравнение. Тригонометрия

Answer 1

(1-cosx+sinx)/cosx=0
cosx≠0 ⇒ x≠π/2+πn, n∈Z
1-cosx+sinx=0
cosx=(1-tg²(x/2))/(1+tg²(x/2))
sinx=(2tg(x/2))/(1+tg²(x/2))
tg(x/2)=m
1-(1-m²)/(1+m²)+2m/(1+m²)=0
(1+m²-1+m²+2m)/(1+m²)=0
(2m²+2m)/(1+m²)=0
2m²+2m=0/:2
m²+m=0
m(m+1)=0
m=0, m=-1
tg(x/2)=0
x/2=πk, k∈Z
x=2πk, k∈Z
tg(x/2)=-1
x/2=3π/4+πk, k∈Z
x=3π/2+2πk, k∈Z
НО: x≠π/2+πn, n∈Z
ПОЭТОМУ:
x=2πk, k∈Z
Ответ: x=2πk, k∈Z