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[tex]\[
\sin \left(90^{\circ}-x\right)=-\frac{\sqrt{3}}{2}
\][/tex]

The value of [tex]\(x\)[/tex] that satisfies the equation is [tex]\(\square\)[/tex] because [tex]\(\square\)[/tex].



Answer :

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].