2
$\begingroup$

Find the number of real solutions on an interval $(0,\pi)$ of this equation $$\sin(14u) - \sin(12u) + 8\sin(u) - \cos(13u) = 4$$ I tried to simplify like this: $2\sin(7u)\cos(7u) - 2\sin(6u)\cos(6u) - 8\sin(u) -\cos(6u)\cos(7u) + \sin(6u)\sin(7u) = 4$ $\cos(7u)(2\sin(7u) - \cos(6u)) + 2\sin(3u)\cos(3u)(\sin(7u) - 2\cos(6u)) - 8\sin(u) = 4$

I could carry on like this until only $\sin(u)$ and $\cos(u)$ remain but it there could be a simpler and faster way. Anyone able to see?

$\endgroup$
3
$\begingroup$

$sin(14u) - sin(12u) + 8sin(u) - cos(13u) = 4 $

$ 2sin(u)cos(13u) + 8sin(u) - cos(13u) - 4 = 0 $

$ (2sin(u)-1)(cos(13u)+4) = 0 $

The only solution is $2sin(u)=1$

$\endgroup$
0
$\begingroup$

This equation can be solved explicitly. Rewrite $\sin 14x - \sin 12x$ as a product. Then solve the equation by factoring.

$\endgroup$
  • $\begingroup$ Ok, thanks i solved it. I get $sinu = 1/2$ at the end $\endgroup$ – diredragon Dec 12 '15 at 10:09
0
$\begingroup$

You have $$2\cos13u\sin u+8\sin u-\cos13u=4$$ $$\Rightarrow(2\cos13u+4)(2\sin u-1)=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.