Question

write the equation of the line(s) that is/are parallel to 10/14 and tangent to y=cos(3x) over [0, pi/2]
I don't know where to start solving this! Any help would be greatly appreciated!

Answer 1

"the line(s) that is/are parallel to 10/14" ---- makes no sense

did you mean that the line(s) has a slope of 10/14 ?
If so, why not reduce that to 5/7 ?

Waiting for clarification.

Answer 2

I'm not actually sure! That's what was said on the question so I'm not sure why it wasn't simplified. Unless "| |" doesn't mean parallel, in which case I completely misunderstood the question.

Answer 3

Well, let's just assume that the lines must have a slope of 10/14

your cosine curve is y = cos(3x)
then dy/dx = -3sin(3x)
that is supposed to be 10/14
-3sin 3x = 10/14
sin 3x = -10/42 = -5/21
3x = 3.382 or 3x = 6.043
x = 1.1273 or x = 2.01143 , but we know 0 ≤ x ≤ π/2, so look for other solutions .....

The period of sin(3x) is 2π/3 , so adding 2π/3 to any answer will yield a new one.
x = 1.1273 + 2π/3 = 3.221 , but that is beyond the domain of x

so our only x = 1.1273
then y = cos (3*1.1273) = -.97124

so our contact point is (1.1273, -.97124)
equation of that tangent:
y + .97124 = (5/7)(x - 1.1273)

looks good here:
www.wolframalpha.com/input/?i=graph+y+%3D+cos%283x%29%2C+y++%3D+%285%2F7%29%28x+-+1.1273%29+-+.97124