# f(x)= (3)(Sinx)(Cosx) Find the equation of the tangent when x=Pi, and when does the rate of the change of the function equal -3/2? ` <br> `

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### 1 Answer

Given `f(x)=3sinxcosx` :

(1) Find the equation of the line tangent to the function when `x=pi` .

The slope of the tangent line is the first derivative evaluated at `x=pi`

`f'(x)=3cos^2x-3sin^2x=3cos2x` (using the extended product property and a trig identity)

`f'(pi)=3cos(2pi)=3` so the slope is 3. The tangent passes through the point `(x,f(x))=(pi,f(pi))` where `f(pi)=3sinpicospi=0` .

**So with slope 3 and the point `(pi,0)` , the equation of the line is:**

`y=3(x-pi)`

(2) When does the rate of change equal -3/2?

Set the first derivative equal to -3/2:

`3cos2x=-3/2`

`cos2x=-1/2`

Then `2x=(2pi)/3 + 2kpi` or `2x=(4pi)/3+2kpi` where `kinZZ` .

**Thus `x=pi/3+kpi` or `(2pi)/3+kpi` **

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