To solve the equation [tex]\(\sin(90^\circ - x) = -\frac{\sqrt{3}}{2}\)[/tex], we can use the trigonometric identity [tex]\(\sin(90^\circ - x) = \cos(x)\)[/tex]. This transforms our equation into:
[tex]\[
\cos(x) = -\frac{\sqrt{3}}{2}
\][/tex]
Next, we need to determine the angles [tex]\(x\)[/tex] where [tex]\(\cos(x) = -\frac{\sqrt{3}}{2}\)[/tex]. Recall that cosine is negative in the second and third quadrants of the unit circle.
First, find the reference angle where [tex]\(\cos(x) = \frac{\sqrt{3}}{2}\)[/tex]:
The reference angle for [tex]\(\cos(x) = \frac{\sqrt{3}}{2}\)[/tex] is [tex]\(30^\circ\)[/tex] because [tex]\(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)[/tex].
Using the reference angle, we can find [tex]\(x\)[/tex] in the appropriate quadrants:
1. Second Quadrant:
- In the second quadrant, the cosine is negative, and the angle is given by:
[tex]\[
x = 180^\circ - \text{reference angle} = 180^\circ - 30^\circ = 150^\circ
\][/tex]
2. Third Quadrant:
- In the third quadrant, the cosine is also negative, and the angle is given by:
[tex]\[
x = 180^\circ + \text{reference angle} = 180^\circ + 30^\circ = 210^\circ
\][/tex]
Therefore, the values of [tex]\(x\)[/tex] that satisfy the equation [tex]\(\sin(90^\circ - x) = -\frac{\sqrt{3}}{2}\)[/tex] are:
[tex]\[
\boxed{150^\circ} \text{ and } \boxed{210^\circ}
\][/tex]
So, the value of [tex]\(x\)[/tex] that satisfies the equation is [tex]\(\boxed{150^\circ \text{ and } 210^\circ} \because\)[/tex].
