Question

Find general solution for $x$

$\sin 8x-\cos 6x=\sqrt{3}(\sin 6x+\cos 8x)$.

Answer 1

Ans. $\frac{n\pi }{7}+\frac{\pi }{12}$ or $n\pi -\frac{3\pi }{4}$
$\sqrt{3}\cos 8x-\sin 8x=-(\cos 6x+\sqrt{3}\sin 6x)$
or $\frac{\sqrt{3}}{2}\cos 8x-\frac{1}{2}\sin 8x$
$=-(\frac{1}{2}\cos 6x+\frac{\sqrt{3}}{2}\sin 6x)$
or $\cos (8x+\frac{\pi }{6})=-\cos (6x-\frac{\pi }{3})=\cos [\pi -(6x-\frac{\pi }{3})]$
$\therefore 8x+\frac{\pi }{6}=2n\pi \pm (\frac{4\pi }{3}-6x)$
$\therefore 14x=2n\pi \frac{4\pi }{3}-\frac{\pi }{6}$
$\therefore x=\frac{n\pi }{7}+\frac{\pi }{2}$ for $+$ive
$2x=2n\pi -\frac{4\pi }{3}-\frac{\pi }{6}$
$\therefore x=n\pi -\frac{3\pi }{4}$ for $-$ive
It can be written as $n\pi -\pi +\pi -\frac{3\pi }{4}$
or $(n-1)\pi +\frac{\pi }{4}$ or $n\pi +\frac{\pi }{4}$.