find the x-coordinate of all points on the curve y=12Xcos(5X)-(30*sqrt3*X^2) + 16, pi/5<X<2pi/5 where the tangent line passes through the point (0,16), (not on the curve)

I have absolutely no idea how to solve this one

If I read your formula correctly,

f(x) = 12x cos(5x)-(30√3 x^2) + 16

The slope at any point on the curve is

f'(x) = 12(cos5x - 5x(sin5x + √3))
So, the equation of the tangent line at x=h is

y-f(h) = f'(h)(x-h)
y-(12h cos(5h)-(30√3 h^2) + 16) = (12(cos5h - 5h(sin5h + √3)))(x-h)

Since this line passes through (0,12), we have

12-(12h cos(5h)-(30√3 h^2) + 16) = (12(cos5h - 5h(sin5h + √3)))(-h)

Hmmm. wolframalpha gets no real solutions for that. Looking at the graph, it does appear that the tangent might pass through (0,12) at about x=0.8

The tangent lines near there are
at x=0.8, y = 20.19-54.66x
at x=0.9, y = 10.58-42.27x

So there should be a solution at about 0.88 or so. Hmmm.

Maybe I made a typo in here somewhere.

To find the x-coordinate of all points on the curve where the tangent line passes through the point (0,16), you need to do the following steps:

1. Differentiate the equation of the curve with respect to x to find the derivative: y' = d/dx[12Xcos(5X) - (30*sqrt3*X^2) + 16].

2. Set the derivative equal to the slope of the tangent line passing through (0,16). Since the point (0,16) does not lie on the curve, you need to find the slope of the line passing through (0,16) and any point on the curve.

3. Let's find that slope. Suppose (a,f(a)) is a point on the curve where f(a) is the y-coordinate. The slope between (0,16) and (a,f(a)) is (f(a) - 16) / (a - 0).

4. Replace f(a) with the value of y on the curve equation: (12acos(5a) - (30*sqrt3*a^2) + 16 - 16) / (a - 0).

5. Set this slope equal to the derivative of the function obtained in step 1: (12acos(5a) - (30*sqrt3*a^2) + 16 - 16) / (a - 0) = y'.

6. Now we have an equation for a. Solve this equation to find the possible values of a that satisfy it. This can be a complex equation to solve analytically, so it might be easier to use a numerical method or a mathematical software program.

7. Once you have the values of a, substitute them into the original curve equation y = 12Xcos(5X) - (30*sqrt3*X^2) + 16 to find the corresponding x-coordinates.

Note: The range given in the problem statement, pi/5 < X < 2pi/5, should be considered while solving the equation for a. Only values of a that fall within this range should be considered as valid solutions.

You may need to use numerical methods or a computer program to get the actual values of x.