Question

Given the unit circle what is the value of x

Answer 1

The equation of the unit circle is

$x^2+y^2=1$$\text{ The point (x,}\frac{3}{4})\text{ lies on the given unit circle.}$

Replace x=x and y=3/4 in the equation, we get

$x^2+(\frac{3}{4})^2=1$

$x^2+\frac{9}{16}=1$

Subtracting 9/4 from both sides, we get

$x^2+\frac{9}{16}-\frac{9}{16}=1-\frac{9}{16}$

$x^2=\frac{16}{16}-\frac{9}{16}$

$x^2=\frac{16-9}{16}$

$x^2=\frac{7}{16}$

Taking square root on both sides, we get

$x=\pm\sqrt[]{\frac{7}{16}}$

$x=\pm\frac{\sqrt[]{7}}{4}$

$x=\frac{\sqrt[]{7}}{4}\text{ or }x=-\frac{\sqrt[]{7}}{4}\text{ }$

Hence the required value of x is

$\text{ }x=-\frac{\sqrt[]{7}}{4}\text{ }$