π 3π nπ 2 f π f 3π 2 π nπ, 3π nπ, n - WebAssign

For the tangent line to be horizontal, f (x) = 0. f (x) = 2sinx + sin2x

f (x) = 2cosx+2sinx cosx = 0 2cosx(1 + sinx) = 0 cosx = 0

or

sinx

=

-1,

so

x

=

2

+2n

or

3 2

+2n,

where

n

is

any

integer.

Now

f

2

= 3 and f

3 2

= -1, so the points on the curve with a horizontal

tangent are

2

+2n,

3

and

3 2

+2n

,

-1

,

where

n

is

any

integer.

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