#### To determine

**To find:** All points on the graph such that the tangent line is horizontal.

#### Answer

The required points such that the tangent line is horizontal are at (π2+2nπ,3)_ and (2nπ−π2,−1)_, where *n* is integer.

#### Explanation

**Given:**

The function is f(x)=2sinx+sin2x .

**Derivative Rule: Difference Rule**

If f(x).and g(x) are both differentiable function, then

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)] (1)

**Calculation:**

Obtain the derivative of f(x).

f′(x)=ddx(f(x))=ddx(2sinx+sin2x)

Apply the difference rule as shown in equation (1),

f′(x)=ddx[2sinx]+ddx[sin2x]=2ddx[sinx]+ddx[1−cos2x2] (∵ sin2x=1−cos2x2)=2(cosx)+ddx(12)−ddx(cos2x2)=2cosx+(0)−(−sin2x⋅22)

Simplify further and obtain the derivative,

f′(x)=2cosx+sin2x=2cosx+2sinxcosx (∵sin2x=2sinxcosx )=2cosx(1+sinx)

Thus, the derivative of f(x)=2sinx+sin2x is f′(x)=2cosx(1+sinx) .

Note that, the function f(x) has a horizontal tangent when f′(x)=0,

f′(x)=02cosx(1+sinx)=0cosx(1+sinx)=0cosx=0 or 1+sinx=0

Here, it is sufficient to find the value of *x* for which cosx=0 and it is expressed as follows,

cosx=cos(π2)

Therefore, the general solution for cosx=0 is x=2nπ±π2.

Substitute x=2nπ+π2 in f(x),

f(2nπ+π2)=2sin(2nπ+π2)+sin2(2nπ+π2)=2(sin2nπcosπ2+cos2nπsinπ2)+(sin2nπcosπ2+cos2nπsinπ2)2 =2(0+1)+(0+1)2=3

Thus, the tangent is horizontal at the point is (2nπ+π2,3).

Substitute x=2nπ−π2 in f(x),

f(2nπ−π2)=2sin(2nπ−π2)+sin2(2nπ−π2)=2(sin2nπcosπ2−cos2nπsinπ2)+(sin2nπcosπ2−cos2nπsinπ2)2 =2(0−1)+(0−1)2=−2+1=−1

Thus, the tangent is horizontal at the point is (2nπ−π2,−1).

Therefore, the required points such that the tangent line is horizontal are at (π2+2nπ,3)_ and (2nπ−π2,−1)_, where *n* is integer.