Answer :
To determine the value of [tex]\( x \)[/tex] for the point [tex]\(\left(x, \frac{\sqrt{3}}{2}\right)\)[/tex] that lies on the unit circle, let's follow these steps:
1. Recall the equation of the unit circle:
The standard equation for the unit circle is given by:
[tex]\[ x^2 + y^2 = 1 \][/tex]
This equation states that any point [tex]\((x, y)\)[/tex] on the unit circle will satisfy this equation.
2. Substitute the given y-coordinate:
We are given the point [tex]\(\left(x, \frac{\sqrt{3}}{2}\right)\)[/tex]. Here, the y-coordinate is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Substitute this value into the unit circle equation:
[tex]\[ x^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \][/tex]
3. Simplify the equation:
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]
Now substitute this back into the equation:
[tex]\[ x^2 + \frac{3}{4} = 1 \][/tex]
4. Solve for [tex]\( x^2 \)[/tex]:
Isolate [tex]\( x^2 \)[/tex] by subtracting [tex]\(\frac{3}{4}\)[/tex] from both sides of the equation:
[tex]\[ x^2 = 1 - \frac{3}{4} \][/tex]
Simplify the right side:
[tex]\[ 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4} \][/tex]
So:
[tex]\[ x^2 = \frac{1}{4} \][/tex]
5. Solve for [tex]\( x \)[/tex]:
To find [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \][/tex]
Thus, [tex]\( x \)[/tex] can be either:
[tex]\[ x = \frac{1}{2} \quad \text{or} \quad x = -\frac{1}{2} \][/tex]
6. Match the solution to the given options:
The values of [tex]\( x \)[/tex] are [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex]. Among the given options:
- [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(-\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\frac{1}{2}\)[/tex]
- [tex]\(\frac{2}{\sqrt{3}}\)[/tex]
The correct value of [tex]\( x \)[/tex] is [tex]\(\frac{1}{2}\)[/tex], which matches option C.
Therefore, the correct answer is:
C. [tex]\(\frac{1}{2}\)[/tex]
1. Recall the equation of the unit circle:
The standard equation for the unit circle is given by:
[tex]\[ x^2 + y^2 = 1 \][/tex]
This equation states that any point [tex]\((x, y)\)[/tex] on the unit circle will satisfy this equation.
2. Substitute the given y-coordinate:
We are given the point [tex]\(\left(x, \frac{\sqrt{3}}{2}\right)\)[/tex]. Here, the y-coordinate is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Substitute this value into the unit circle equation:
[tex]\[ x^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \][/tex]
3. Simplify the equation:
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]
Now substitute this back into the equation:
[tex]\[ x^2 + \frac{3}{4} = 1 \][/tex]
4. Solve for [tex]\( x^2 \)[/tex]:
Isolate [tex]\( x^2 \)[/tex] by subtracting [tex]\(\frac{3}{4}\)[/tex] from both sides of the equation:
[tex]\[ x^2 = 1 - \frac{3}{4} \][/tex]
Simplify the right side:
[tex]\[ 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4} \][/tex]
So:
[tex]\[ x^2 = \frac{1}{4} \][/tex]
5. Solve for [tex]\( x \)[/tex]:
To find [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[ x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \][/tex]
Thus, [tex]\( x \)[/tex] can be either:
[tex]\[ x = \frac{1}{2} \quad \text{or} \quad x = -\frac{1}{2} \][/tex]
6. Match the solution to the given options:
The values of [tex]\( x \)[/tex] are [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex]. Among the given options:
- [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(-\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\frac{1}{2}\)[/tex]
- [tex]\(\frac{2}{\sqrt{3}}\)[/tex]
The correct value of [tex]\( x \)[/tex] is [tex]\(\frac{1}{2}\)[/tex], which matches option C.
Therefore, the correct answer is:
C. [tex]\(\frac{1}{2}\)[/tex]