find the exact values of s in the given interval that satisfy the given condition. 0, 2π), cos s =…

find the exact values of s in the given interval that satisfy the given condition. 0, 2π), cos s = \\frac{\\sqrt{3}}{2} a. \\frac{2π}{3}, \\frac{4π}{3} b. \\frac{π}{2}, \\frac{3π}{2} c. \\frac{π}{6}, \\frac{11π}{6} d. \\frac{3π}{4}, \\frac{5π}{4}

find the exact values of s in the given interval that satisfy the given condition. 0, 2π), cos s = \\frac{\\sqrt{3}}{2} a. \\frac{2π}{3}, \\frac{4π}{3} b. \\frac{π}{2}, \\frac{3π}{2} c. \\frac{π}{6}, \\frac{11π}{6} d. \\frac{3π}{4}, \\frac{5π}{4}

Answer

Explanation:

Step1: Recall the unit - circle values

We know that (\cos s=\frac{\sqrt{3}}{2}). On the unit circle, (\cos\theta = x) - coordinate of the point ((x,y)) corresponding to the angle (\theta). We know that (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}).

Step2: Use the property of cosine function

The cosine function has the property (\cos(2\pi - \alpha)=\cos\alpha). Let (\alpha=\frac{\pi}{6}), then (\cos(2\pi-\frac{\pi}{6})=\cos\frac{11\pi}{6}) and (\cos\frac{11\pi}{6}=\frac{\sqrt{3}}{2}) since (2\pi-\frac{\pi}{6}=\frac{12\pi - \pi}{6}=\frac{11\pi}{6}) and for (s\in[0,2\pi)), when (\cos s=\frac{\sqrt{3}}{2}), (s = \frac{\pi}{6}) or (s=\frac{11\pi}{6})

Answer:

C. (\frac{\pi}{6},\frac{11\pi}{6})