Certainly! Let's solve the problem step-by-step.
Given the point [tex]\((x, \frac{\sqrt{3}}{2})\)[/tex] is on the unit circle, we need to find the possible value(s) of [tex]\(x\)[/tex].
1. Unit Circle Equation:
[tex]\[
x^2 + y^2 = 1
\][/tex]
Here, [tex]\(y = \frac{\sqrt{3}}{2}\)[/tex].
2. Substitute [tex]\(y\)[/tex] into the equation:
[tex]\[
x^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 1
\][/tex]
3. Calculate [tex]\(\left(\frac{\sqrt{3}}{2}\right)^2\)[/tex]:
[tex]\[
\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{2^2} = \frac{3}{4}
\][/tex]
4. Substitute [tex]\(\frac{3}{4}\)[/tex] back into the equation:
[tex]\[
x^2 + \frac{3}{4} = 1
\][/tex]
5. Solve for [tex]\(x^2\)[/tex]:
[tex]\[
x^2 = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4}
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Since [tex]\(x^2 = \frac{1}{4}\)[/tex],
[tex]\[
x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}
\][/tex]
Thus, the possible values for [tex]\(x\)[/tex] are [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex].
Now, let's evaluate the given options:
- A. [tex]\(\frac{2}{\sqrt{3}}\)[/tex]: This is incorrect because [tex]\(\frac{2}{\sqrt{3}}\)[/tex] does not equate to [tex]\(\frac{1}{2}\)[/tex] or [tex]\(-\frac{1}{2}\)[/tex].
- B. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]: This is incorrect since it doesn't match [tex]\(\frac{1}{2}\)[/tex] or [tex]\(-\frac{1}{2}\)[/tex], and would not satisfy the original equation [tex]\(x^2 + y^2 = 1\)[/tex].
- C. [tex]\(-\frac{\sqrt{3}}{2}\)[/tex]: Similarly, this is incorrect for not equating [tex]\(\frac{1}{2}\)[/tex] or [tex]\(-\frac{1}{2}\)[/tex], and does not fit the unit circle criteria with the given [tex]\(y\)[/tex].
- D. [tex]\(\frac{1}{2}\)[/tex]: This is correct since [tex]\(\frac{1}{2}\)[/tex] matches one of the solutions we found.
Therefore, the correct option is:
[tex]\[
\boxed{\frac{1}{2}}
\][/tex]
