How do you determine at which the graph of the function #y=sqrt(3x)+2cosx# has a horizontal tangent line?

To determine where the graph of the function y = sqrt(3x) + 2cos(x) has a horizontal tangent line, we need to find the points where the derivative of the function is equal to zero.

First, we find the derivative of the function by applying the chain rule and product rule:

dy/dx = (1/2)*(3x)^(-1/2)*3 + (-2)*sin(x)

Next, we set the derivative equal to zero and solve for x:

(1/2)*(3x)^(-1/2)*3 + (-2)*sin(x) = 0

Simplifying the equation, we get:

(3/2)(3x)^(-1/2) = 2sin(x)

To solve this equation, we can square both sides:

(9/4)(3x)^(-1) = 4sin^2(x)

Further simplifying, we have:

(9/4)(1/(3x)) = 4sin^2(x)

Cross-multiplying, we get:

9 = 16sin^2(x)(3x)

Dividing both sides by 16*sin^2(x), we obtain:

9/(16*sin^2(x)) = 3x

Finally, we solve for x:

x = 3/(16*sin^2(x))

By solving this equation numerically or graphically, we can find the values of x where the graph of the function has a horizontal tangent line.