Question

Seja g(x)= x^2 + xcosbeta+ senbeta. Se g (x)=0 e beta=270° então, calcule o valor de X.

Answer 1

g(x)=x²+xcosb+senb

g(x)=0

b=270°=3π/2

x²+xcos(3π/2)+sen(3π/2)=0

x²-1=0

x²=1

x=+-1

S:{-1,1}.