Find the solution to the initial value problem $$f '(x)=4(\cos u-\sin2 u); f(\frac{ \pi }{6 })=0$$

Question

Find the solution to the initial value problem $$f '(x)=4(\cos u-\sin2 u);   f(\frac{ \pi }{6 })=0$$

anonymous · Answer

So far I have $$f(x)=\sin 4u+2\cos 2u+C$$but when I try to plug in pi/6 for u I'm not getting the correct answer for C which should be 3 (according to the back of the book)

anonymous · Answer

-3 that is

anonymous · Answer

@ZeHanz

zehanz · Answer

The factor 4 can be ignored. Put it as first factor in f(x):

$f(x)=4( \sin u +\frac{1}{2}\cos2u)$. Oh, now I can see your error:

$f(x)=4\sin x + 2\cos x +C$. You typed sin4x...

zehanz · Answer

This function will give you C=-3, if you set $f(\frac{\pi}{6})=0$.

anonymous · Answer

oh you're right

anonymous · Answer

LOL thanks alot

zehanz · Answer

YW!
I make these errors myself all the time!

anonymous · Answer

Good thing you caught this one. I was really confused