Respuesta :

naǫ
 The equation of the unit circle:
[tex]x^2+y^2=1[/tex]

[tex](x,y)=(x, \frac{\sqrt{3}}{3}) \\ \\ x^2+(\frac{\sqrt{3}}{3})^2=1 \\ x^2+\frac{3}{9}=1 \\ x^2+\frac{1}{3}=1 \\ x^2=1-\frac{1}{3} \\ x^2=\frac{3}{3}-\frac{1}{3} \\ x^2=\frac{2}{3} \\ x=\pm \sqrt{\frac{2}{3}} \\ x=\pm \frac{\sqrt{2}}{\sqrt{3}} \\ x=\pm \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\ x=\pm \frac{\sqrt{6}}{3} [/tex]

x is equal to [tex]-\frac{\sqrt{6}}{3}[/tex] or [tex]\frac{\sqrt{6}}{3}[/tex].

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