## anonymous one year ago if the point(x, sqrt 3/2) is on the unit circle, what is x

1. jim_thompson5910

The point is $\Large \left(x, \frac{\sqrt{3}}{2}\right)$ right?

2. anonymous

yes

3. jim_thompson5910

Notice how that point is the form (x,y) where $\Large y = \frac{\sqrt{3}}{2}$ Do you agree with this statement?

4. anonymous

yes

5. jim_thompson5910

Now plug this y value into the unit circle equation and solve for x $\Large x^2+y^2 = 1$ $\Large x^2+\left( \frac{\sqrt{3}}{2}\right)^2 = 1$ $\Large x^2+ \frac{\left(\sqrt{3}\right)^2}{2^2} = 1$ $\Large x^2+ \frac{3}{4} = 1$ Do you see how to solve for x from here?

6. anonymous

no

7. jim_thompson5910

how would you move the 3/4 to the other side? what must you do to both sides?

8. jim_thompson5910

hint: you need to undo the addition

9. anonymous

subtract 3/4 from each side

10. jim_thompson5910

yes, that gives you $\Large x^2+ \frac{3}{4} = 1$ $\Large x^2+ \frac{3}{4}\color{red}{- \frac{3}{4}} = 1\color{red}{- \frac{3}{4}}$ $\Large x^2 = 1- \frac{3}{4}$ $\Large x^2 = \frac{4}{4}- \frac{3}{4}$ $\Large x^2 = \frac{4-3}{4}$ $\Large x^2 = \frac{1}{4}$

11. jim_thompson5910

Now if $\Large x^2 = \frac{1}{4}$ then what must x be? How would you isolate x?

12. anonymous

1/2

13. jim_thompson5910

x = 1/2 is one solution what is the other?

14. anonymous

i don't know

15. jim_thompson5910

hint: take a number like 10 and square it to get 100. You can also square -10 to get 100 as well since (-10)^2 = (-10)*(-10) = 100 so that's why the solutions to x^2 = 100 are x = -10 or x = 10