Răspuns:
Explicație pas cu pas:
[tex]2sin^{2}\frac{x}{2}= 1-cosx,~~obtinem~ecuatia~sinx+cosx=1-cosx,~deci~sinx+2cosx=1.~~Aplicam~formulele~sinx=\dfrac{2tg\frac{x}{2}}{1+tg^{2}\frac{x}{2} }~~si~~cosx= \dfrac{1-tg^{2}\frac{x}{2} }{1+tg^{2}\frac{x}{2} }.\\Obtinem:~~\dfrac{2tg\frac{x}{2}}{1+tg^{2}\frac{x}{2} }+2*\dfrac{1-tg^{2}\frac{x}{2}}{1+tg^{2}\frac{x}{2} }=1.~~inmultim~la~1+tg^{2}\frac{x}{2},~obtinem,\\2tg\frac{x}{2}+2(1-tg^{2}\frac{x}{2})=1+tg^{2}\frac{x}{2},~notam ~tg\frac{x}{2}=a.[/tex]
Obtinem 2a+2(1-a²)=1+a², ⇒2a+2-2a²-a²-1=0, ⇒-3a²+2a+1=0, ⇒3a²-2a-1=0
Δ=4+12=16, deci a1=(2-4)/6=-1/3 si a2=(2+4)/6=1
Deci tg(x/2)=-1/3, ⇒x/2=-arctg(1/3)+kπ, ⇒x=-2arctg(1/3)+2kπ, unde k∈Z.
Sau tg(x/2)=1, ⇒x/2=arctg1+kπ, ⇒x/2=π/4 +kπ, ⇒x=π/2 + 2kπ, unde k∈Z.
Raspuns: S={-2arctg(1/3)+2kπ; π/2 + 2kπ | k∈Z}.
