Question

Find the general solutions of each of the following equations:
(i) $\cos 4x=\cos 2x$
(ii) $\cos 3x=\sin 2x$
(iii) $\sin mx+\sin nx=0$

Answer 1

(i)

$\cos 4x=\cos 2x$

$\cos 4x-\cos 2x=0$

$-2\sin \left(\frac{4x+2x}{2}\right)\sin \left(\frac{4x-2x}{2}\right)$

$=-2\sin 3x\sin x=0$

$\sin 3x=0,\sin x=0$

General solution of $\sin 3x$

$3x=n\pi$

$x=\frac{n\pi }{3}$

General solution of $\sin x$

$x=n\pi$

(ii)

$\cos 3x=\sin 2x$

$\cos 3x=\cos (\frac{\pi }{2}-2x)$

$\pm 3x=\frac{\pi }{2}-2x$

(a) $\frac{\pi }{2}=5x$

$x=\frac{\pi }{10}$

(b) $\frac{\pi }{2}=-x$

$x=-\frac{\pi }{2}$

$\cos 3x=\cos (\frac{\pi }{2}-2x-2\pi )$

$\pm 3x=\frac{\pi }{2}-2x-2\pi$

(c) $\frac{\pi }{2}-2\pi =5x$

$x=\frac{3}{10}\pi$

(d) $\frac{\pi }{2}-2\pi -x$

$x=\frac{3}{2}\pi$

(iii)

$\sin mx+\sin nx=0$

$2\sin \left(\frac{mx+nx}{2}\right)\cos \left(\frac{mx-nx}{2}\right)=0$

$\sin \left(\frac{mx+nx}{2}\right)=0,\cos \left(\frac{mx-nx}{2}\right)=0$

$\left(\frac{m+n}{2}\right)x=px,\left(\frac{m-n}{2}\right)x=(2q+1)\frac{\pi }{2}$

$x=\frac{2px}{m+n},x=\frac{(2q+1)\pi }{m-n}$

p and q are integers.