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,}(3)/(4))\text{ lies on the given unit circle.}`

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

`x^2+((3)/(4))^2=1`

`x^2+(9)/(16)=1`

Subtracting 9/4 from both sides, we get

`x^2+(9)/(16)-(9)/(16)=1-(9)/(16)`

`x^2=(16)/(16)-(9)/(16)`

`x^2=(16-9)/(16)`

`x^2=(7)/(16)`

Taking square root on both sides, we get

`x=\pm\sqrt[]{(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{ }`