for what values of ( x ) does ( f(x)=2 cos x-sqrt{3} cdot x ) have a horizontal tangent line?\ngive all…

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

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

Answer

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})