Given the function f with the restricted domain, find the points at which the tangent line is horizontal.

f(x)=2sec(x)-tan(x)
Domain: [-pi/2,pi/2]

* I got (pi/6, square root of 3), but I am not sure if that is right, or if there is another point.

To find points at which the tangent line is horizontal for a given function, you need to find the values of x where the derivative of the function is equal to zero. In other words, you need to find the values of x where the slope of the tangent line is zero.

To approach this problem, let's first find the derivative of the function f(x). The derivative gives us the slope of the tangent line at any given point on the curve. Then, we can set the derivative equal to zero and solve for x.

1. Start with the function f(x) = 2sec(x) - tan(x) and find its derivative.

To find the derivative, we can use the rules for derivatives of trigonometric functions. The derivative of sec(x) is sec(x)tan(x), and the derivative of tan(x) is sec^2(x).

So, f'(x) = 2sec(x)tan(x) - sec^2(x).

2. Now, set f'(x) (the derivative) equal to zero and solve for x.

2sec(x)tan(x) - sec^2(x) = 0

To simplify this equation, let's factor out a common term:

sec(x)(2tan(x) - sec(x)) = 0

Now, we have either sec(x) = 0 OR 2tan(x) - sec(x) = 0

For the domain [-pi/2, pi/2], sec(x) is never equal to zero. So, the only possible solution is 2tan(x) - sec(x) = 0.

2tan(x) - sec(x) = 0

3. Solve the equation 2tan(x) - sec(x) = 0 for x.

To solve this equation, we can rewrite it as:

2sin(x)/cos(x) - 1/cos(x) = 0

Multiplying through by cos(x), we get:

2sin(x) - 1 = 0

Solving for sin(x), we have:

sin(x) = 1/2

From the equation sin(x) = 1/2, we know that x must be pi/6 or 5pi/6. However, since the given domain is [-pi/2, pi/2], only pi/6 is a valid solution within the given domain.

Therefore, the only point at which the tangent line is horizontal for the function f(x) = 2sec(x) - tan(x), with the domain [-pi/2, pi/2], is (pi/6, sqrt(3)).

So your answer of (pi/6, sqrt(3)) is correct.