Question

Please help me with this math problem (:

Answer 1

`{x}^(2) + ( ( √(3) )/(2) ) {}^(2) = 1 \\ {x}^(2) = (1)/(4) \\ x = (1)/(2) \: \: or( (1)/(2) ) \\ x > 0 \\ therefore \: \: x = (1)/(2)`

Answer 2

`x=-(1)/(2)`

Explanation:

equation of a circle:
`(x - h)^2+(y-k)^2=r^2`

where (h, k) = center and r = radius

From inspection, we can see that the center of the circle = (0, 0)

and the radius = 1

`\implies (x - 0)^2+(y-0)^2=1^2`

`\implies x^2+y^2=1`

Substitute
`y=(√(3))/(2)` into
`x^2+y^2=1` and solve for
`x`:

`\implies x^2+((√(3))/(2))^2=1`

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

`\implies x^2=(1)/(4)`

`\implies x=\pm \sqrt{(1)/(4)}`

`\implies x=\pm(1)/(2)`

As point P is in quadrant 2,
`x` is negative, so
`x=-(1)/(2)`