Question

find the x-coordinates of all points on the graph of: f(x)=6sinx-sin^2x
at which the tangent line is horizontal.
I get to deriving the function above then Im lost!

Answer 1

*f(x)=6sinx-sin^2(x)

Answer 2

once you get the derivative, set it equal to zero and solve

Answer 3

my answers are in terms of pi/2, pi/3, pi/4 with options for 2npi or just npi

Answer 4

\[\frac{d}{dx}6\sin(x)-\sin^2(x)=6\cos(x)-2\sin(x)\cos(x)\]
set
\[6\cos(x)-2\sin(x)\cos(x)=0\] then
\[2\cos(x)(3-2\sin(x))=0\]
so
\[\cos(x)=0\] or
\[2\sin(x)=3\]
\[\sin(x)=\frac{3}{2}\] and then solve for x

Answer 5

oops i made a mistake in factoring i think

Answer 6

\[6\cos(x)-2\sin(x)\cos(x)=0\]
\[2\cos(x)(3-\sin(x))=0\]

Answer 7

so \[\cos(x)=0\] or
\[\sin(x)=3\] which is not possible

Answer 8

therefore
\[\cos(x)=0,x=\frac{\pi}{2}+n\pi, n\in \mathbb Z\]

Answer 9

Thanks for walking me through it, i really appreciate it!