Find the x-coordinates of all points on the curve $\displaystyle y = sin2x - 2sinx$ at which the tangent line is horizontal.
I don't know what to do and where to begin.
for the first part we use the chain rule:
$\displaystyle \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$
for $\displaystyle \sin 2x$: $\displaystyle f(x) = \sin x$ and $\displaystyle g(x) = 2x$
the other piece is trivial. $\displaystyle \frac d{dx} \sin x = \cos x$ so $\displaystyle \frac d{dx} -2 \sin x = -2 \frac d{dx} \sin x = -2 \cos x$
are you taking me for a ride? or are you really trying? *suspicious smiley (we still need one of these)*
$\displaystyle 2 \cos^2 x - \cos x - 1 = 0$ ....this is a quadratic, see if you can factorize
$\displaystyle \Rightarrow (2 \cos x + 1)(\cos x - 1) = 0$
$\displaystyle \Rightarrow 2 \cos x + 1 = 0 \mbox{ or } \cos x - 1 = 0$
$\displaystyle \Rightarrow \cos x = - \frac 12 \mbox{ or } \cos x = 1$
what angles does this happen for?