Question

Solve $\sin 5x.\cos 3x=\sin 6x.\cos 2x$. then $\frac{n\pi }{2},n\in I$ or $n\pi \pm \frac{\pi }{6},n\in I$
If true then enter $1$ and if false then enter $0$

Answer 1

$\because \sin 5x.\cos 3x=\sin 6x.\cos 2x$
$\Rightarrow 2\sin 5x.\cos 3x=2\sin 6x.\cos 2x$
$\Rightarrow \sin 8x+\sin 2x=\sin 8x+\sin 4x$
$\Rightarrow \sin 4x-\sin 2x=0$
$\Rightarrow 2\sin 2x.\cos 2x-\sin 2x=0$
$\Rightarrow \sin 2x(2\cos 2x-1)=0$
$\Rightarrow \sin 2x=0$ or $2cos2x-1=0$
$\Rightarrow 2x=n\pi \in I$ or $\cos 2x=\frac{1}{2}$
$\Rightarrow x=\frac{n\pi }{2},n\in I$
$\Rightarrow x=n\pi \pm \frac{\pi }{6},n\pi I$
$\therefore$ Solution of given equation is $\frac{n\pi }{2},n\in I$ or $n\pi \pm \frac{\pi }{6},n\in I$