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

so

3 = 2+2sinx

=

1 = 2sinx2

=

1/2 = sinx

=

sin^-1(1/2) = x

= 0.524

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 ?