Question

Resolver, na variável x, a equação x² − 2x + sen² α = 0

Answer 1

Resposta:

x=1-cosα ou x=1+cosα

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-2x+sen^2\alpha=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-2~e~c=sen^2\alpha\\\\\Delta=(b)^{2}-4(a)(c)=(-2)^{2}-4(1)(sen^2\alpha)=4-4sen^2\alpha=4(1-sen^2\alpha)=4cos^2\alpha\\\\\sqrt{\Delta}=\sqrt{4cos^2\alpha}=\sqrt{4}.\sqrt{cos^2\alpha}=2cos\alpha$

$\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-2)-2cos\alpha}{2(1)}=\frac{2-2cos\alpha}{2}=\frac{2(1-cos\alpha)}{2}=1-cos\alpha\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-2)+2cos\alpha}{2(1)}=\frac{2+2cos\alpha}{2}=\frac{2(1+cos\alpha)}{2}=1+cos\alpha$

Relação fundamental da trigonometria:

cos²x+sen²x=1

cos²x=1-sen²x