# Real solutions to trigonometric equation on an interval $(0, \pi)$

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?

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

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

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

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

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

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

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