Hello, I am trying to solve the following problem:
At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.
My attempt is:
rewrite 3x-y=5 as y=3x-5
differentiated to 3
Differentiate y=2(x-cosx) = 2(1+sinx) = 2 + 2sinx
3 = 2+2sinx
1 = 2sinx2
1/2 = sinx
sin^-1(1/2) = x
plugging back into original curve formula gives a y value of -0.684
so my final answer is "the tangent is parallel to the line 3x-y=5 at the point .524(x), -0.684(y)."
I am unsure if I am correct. Could someone please cast their eyes over this and let me know if I am on the right track ?