Question

for what values of ( x ) does ( f(x)=2 cos x-sqrt{3} cdot x ) have a horizontal tangent line?
give all answers in the interval ( 0 leq x<2 pi ). give your answers as exact values, separated by commas.
( x= )

Explanation:

Step1: Find the derivative of \( f(x) \)

The derivative of \( f(x)=2\cos x-\sqrt{3}x \) is \( f^\prime(x)=-2\sin x-\sqrt{3} \).

Step2: Set the derivative equal to zero

Since a horizontal tangent line occurs when \( f^\prime(x) = 0 \), we set \( -2\sin x-\sqrt{3}=0 \).
This gives \( \sin x=-\frac{\sqrt{3}}{2} \).

Step3: Solve for \( x \) in the interval \( 0\leq x < 2\pi \)

In the interval \( 0\leq x < 2\pi \), the solutions of \( \sin x =-\frac{\sqrt{3}}{2} \) are \( x=\frac{4\pi}{3},\frac{5\pi}{3} \).

Answer:

\(\frac{4\pi}{3},\frac{5\pi}{3}\)