Hello, and thanks in advance for the help.

The problem I'm given in this review for my upcoming exam is to find the equation of the tangent line to x = y - cos(y) at the point (-1, 0).

I've tried several things and am just hoping one of the local math gurus could make sure I've done my steps correctly :-).

I know I need to find y' at the point (-1, 0) because it's the slope of the tangent line. My first attempt at finding y' yielded y' = 1 / sin(y), where inputting 0 for y caused 1 / 0.

However, my second attempt is much more promising if it leads where I think, and I'll just list the steps below:

x = y - cos(y)

x - y = -cos(y)

1 - y'(y^0) = sin(y) [took the derivative of both sides, and y^0 = 1)

y' = 1 + sin(y)

Does this mean that at the point (-1, 0), the slope is 1 + sin(0) or m = 1?

If so, the equation for the tangent line would be y = x + 1, right?

Thanks again, this forum is a great resource!