# How do I find the coordinates of the points on the curve y = (2-cos(x)) / sin(x) , in the range 0<x<2pi , where the gradient is 3/8.

### 1 Answer | Add Yours

(2-cos(x))/sinx

==> y' =sinx -(2-cosx)cosx)/sin^2 x

==> (sin^x -2cosx+cos^2 x)/sin^2 x

==> (1-2cosx)/sin^2 x

==> (1-2cosx)/sin^2 x = 3/8

==> 8(1-2cosx) = 3sin^2 x

==> 8 - 16cosx = 3-3cos^2 x

==> 3cos^2 x -16cosx +5 = 0

==. (3cosx -1)(cosx -5) = 0

==> (3cosx-1)= 0 ==> cosx = 1/3 ==> x=70.5

==> cosx-5 = 0 ==> cosx = 5 ( Invalid)

==> x= 70.5 degrees.

==> y= (2-cosx)/sinx = (2- 1/3)/ (0.94)= 1.77

**==>Then the point (x,y)= (70.5, 1.77) such that the gradient is 3/8.**