The question can be represented by:
where
[tex]y=\frac{\sqrt[]{3}}{2}[/tex]
Note that the Hypotenuse is 1 because a unit circle has a radius of 1.
Considering the triangle from the diagram above, we can find x using the Pythagorean Theorem:
[tex]\begin{gathered} 1^2=x^2+y^2 \\ 1=x^2+(\frac{\sqrt[]{3}}{2})^2 \\ 1=x^2+\frac{3}{4} \end{gathered}[/tex]
Solving for x, we have
[tex]\begin{gathered} x^2=1-\frac{3}{4} \\ x^2=\frac{1}{4} \\ x=\sqrt[]{\frac{1}{4}} \\ x=\frac{1}{2} \end{gathered}[/tex]
The value of x is 1/2.
