Find the equation of the tangent to the curve (sinusoidal function)

[Saterial]

Homework Statement

Find the equation of the tangent to the curve y=2cos^3x at x=pi/3

Homework Equations

The Attempt at a Solution

y=2cos^3x
dy/dx=-6sinxcos^2x
0=-6sinxcos^2x

set x = pi/3 and solve for the derivative, plug the answer into y=2cos^3x

where do I go from here?

Do I take that value of the slope and use the points found by pi/3 and use m=y2-y1/x2-x1?

Thanks

 

[HallsofIvy]

The equation of a line through the point [itex](x_0, y_0)[/itex] with slope "m" is
[itex]y= m(x- x_0)+ y_0[/itex].

You can derive that from the formula
[tex]m= \frac{y- y_0}{x- x_0}[/tex]

 

[Mentallic]

Homework Statement

Find the equation of the tangent to the curve y=2cos^3x at x=pi/3

Homework Equations

The Attempt at a Solution

y=2cos^3x
dy/dx=-6sinxcos^2x
0=-6sinxcos^2x

set x = pi/3 and solve for the derivative, plug the answer into y=2cos^3x

where do I go from here?

Do I take that value of the slope and use the points found by pi/3 and use m=y2-y1/x2-x1?

Thanks

You should read dy/dx as being the gradient. If you want to know the gradient at x=1, then you find dy/dx and then substitute x=1 into that equation. For example, dy/dx=10x, at x=1 dy/dx=10 so the gradient at that point is 10. If you want to know where the gradient is equal to 2, then substitute dy/dx=2 and solve for x, thus 2=10x, x=1/5 so at x=1/5 you'll have a gradient of 2.
You substituted dy/dx=0 which means you're looking for the values of x where the gradient is 0 (or a turning point in other words) but that's not what you're looking for - you're looking for the value of dy/dx at x=pi/3.