Question

Rezolvati ecuatia
sin2x+sinx+cosx=1

Answer 1

folosim regula de substitutie universala

sin(x)=2tg(x/2) / (1+tg^2(x/2)

cos(x)=(1-tg^2(x/2))/(1+tg^2(x/2))

vom nota tg(x/2) cu t

sin2x=2sinxcosx

2*(2t/(1+t^2))*((1-t^2)/(1+t^2))+(2t/(1+t^2))+((1-t^2)/(1+t^2))=1

4t(1-t^2)/(1+t^2)^2+(2t/(1+t^2))+((1-t^2)/(1+t^2))=1

(4t-4t^3)/(1+t^2)^2+(2t/(1+t^2))+((1-t^2)/(1+t^2))-1=0

(4t-4t^3+2t(1+t^2)+(1+t^2)+(1+t^2)(1-t^2)-(1-t^2)^2)/(1+t^2)^2=0

(4t - 4t^3 + 2t +2t^3 + 1 - t^4 - (1+2t^2+t^4))/(1+t^2)^2=0

(6t-2t^3-2t^4-2t^2)/(1+t^2)^2=0

(6t-2t^3-2t^4-2t^2)=0

2t(-t^3+t^2-2t^2+2t-3t+3)=0

2t(-t^2(t-1)-2t(t-1)-3(t-1))=0

2t(-(t-1)(t^2+t+3))=0

t(t-1)(t^2+2t+3)=0

t^2+2t+3 nu are solutii pe R => t poate fi 0 sau 1

t este tg(x/2)

tg(x/2)=0 deci x=2k*pi unde k apartine lui Z

tg(x/2)=1 deci x=pi/2 + 2l*pi unde k apartine lui Z