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Answer:
[tex]\theta = 30^\circ, 330^\circ[/tex]
[tex]\theta = \frac{\pi}{6}, \frac{11\pi}{6}[/tex]
Step-by-step explanation:
Given [Missing from the question]
Equation:
[tex]cos\theta = \frac{\sqrt 3}{2}[/tex]
Interval:
[tex]0 \le \theta \le 360[/tex]
[tex]0 \le \theta \le 2\pi[/tex]
Required
Determine the values of [tex]\theta[/tex]
The given expression:
[tex]cos\theta = \frac{\sqrt 3}{2}[/tex]
... shows that the value of [tex]\theta[/tex] is positive
The cosine of an angle has positive values in the first and the fourth quadrants.
So, we have:
[tex]cos\theta = \frac{\sqrt 3}{2}[/tex]
Take arccos of both sides
[tex]\theta = cos^{-1}(\frac{\sqrt 3}{2})[/tex]
[tex]\theta = 30[/tex] --- In the first quadrant
In the fourth quadrant, the value is:
[tex]\theta = 360 -30[/tex]
[tex]\theta = 330[/tex]
So, the values of [tex]\theta[/tex] in degrees are:
[tex]\theta = 30^\circ, 330^\circ[/tex]
Convert to radians (Multiply both angles by [tex]\pi/180[/tex])
So, we have:
[tex]\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}[/tex]
[tex]\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}[/tex]
[tex]\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}[/tex]
[tex]\theta = \frac{\pi}{6}, \frac{11\pi}{6}[/tex]