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Find the point on the curve y=cosx-1,...

Find the point on the curve `y=cosx-1,\ \ x in [pi/2,(3pi)/2]` at which the tangent is parallel to the x-axis.

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Given, `y=cosx-1,\ \ x in [pi/2,(3pi)/2]`

`(dy)/(dx)=-sinx`

`(dy)/(dx)=0`

`-sin x=0` (tangent parallel to `x` axis)

`implies sin x=0`

`x=pi`

Hence,

`y=cos pi-1=(-1)-1=-2`

...
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