Finding the equation of the tangent

Given there are distinct points on the curve y= 3sinx-x which have a common tangent, find 2 such tangents, their equations and points of contact.

I get the 1st differential (wrt x) to be 3cosx -1. But apparently the answer is:

when x=X
y= (3cosX - 1)x + 3(sinX - XcosX).

How?

You haven't gone far enough.

Next step: the equation of the straight line is y = mx + c.

I did, I just didn't get the same equation as the answer states.

I got

y-Y=(3cosx - 1)x -X(3cosx - 1)

Where X= x coordinate and Y=y coordinate

I don't know why you have substituted the gradient twice. Don't get wedded to the y-y1 = m(x - x1) formula if you don't understand it properly.

How do you mean?

I just expanded the right hand side of the y-y1 = m(x-x1) equation?

Is that wrong?

Oh I see. You are making this far too difficult.

What point are you going to substitute if you follow this path?

Tbh I'm not sure where to go from there because my solution doesn't look a lot like the text book solution I posted. Any ideas on how to solve this?

I told you how to do it in post 2.

y = mx + c

then you have an equation for y to substitute to complete the question.

where does the sinx in the answer come from though? I get c as -1.

Given there are distinct points on the curve y= 3sinx-x which have a common tangent, find 2 such tangents, their equations and points of contact.

I get the 1st differential (wrt x) to be 3cosx -1. But apparently the answer is:

when x=X
y= (3cosX - 1)x + 3(sinX - XcosX).

How?

Eq of tangent will be y = mx+c as MrM points out
At the point where x = X the value of m will be 3cosX - 1
Now you need to find the value of the constant c.

You know that the tangent where x=X must intersect the original curve where x=X, so ...

Eq of tangent will be y = mx+c as MrM points out
At the point where x = X the value of m will be 3cosX - 1
Now you need to find the value of the constant c.

You know that the tangent where x=X must intersect the original curve where x=X, so ...

just revisited this and think I've forgotten how to get the c, can you please help? the tangent intersects the original curve where x=x1 so how do we get c to be 3(sinx1-x1-x1cosx1) ?

At the point (X, 3sinX - X) we found the gradient of the tangent to be m = 3cos X - 1.

Therefore our tangent line is y = (3cosX - 1)x + c

So now we plug in our coordinates of the point: x=X, y=3sinX - X, rearrange and solve for c.