# Find the general solution of each of the following equations:

Question:

Find the general solution of each of the following equations:

(i) $\cos 3 x=\cos 2 x$

(ii) $\cos 5 x=\sin 3 x$

(iii) $\cos \mathrm{m} \mathrm{x}=\sin \mathrm{n} \mathrm{x}$

Solution:

To Find: General solution.

(i) Given: $\cos 3 x=\cos 2 x \Rightarrow \cos 3 x-\cos 2 x=0 \Rightarrow-2 \sin \frac{(5 x)}{2} \sin \frac{(x)}{2}=0$

[NOTE: $\left.\cos C-\cos D=-2 \sin \frac{(C+D)}{2} \sin \frac{(C-D)}{2}\right]$

So, $\sin \frac{(5 x)}{2}=0$ or $\sin \frac{(x)}{2}=0$

Formula used: $\sin \theta=0 \Rightarrow \theta=n \pi, n \in I$

$\frac{(5 x)}{2}=n \pi$ or $\frac{(x)}{2}=m \pi$ where $n, m \in I$

$x=2 \mathrm{n} \pi / 5$ or $x=2 \mathrm{~m} \pi$ where $n, m \in I$

So general solution is $x=2 \mathrm{n} \pi / 5$ or $x=2 \mathrm{~m} \pi$ where $n, m \in I$

(ii) Given: $\cos 5 x=\sin 3 x \Rightarrow \cos 5 x=\cos \left(\frac{\pi}{2}-3 x\right)$

Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in I$

By using the above formula, we have

$5 x=2 n \pi+\left(\frac{\pi}{2}-3 x\right)$ or $5 x=2 n \pi-\left(\frac{\pi}{2}-3 x\right)$

$8 x=2 n \pi+\frac{\pi}{2}$ or $2 x=2 n \pi-\frac{\pi}{2}$

$x=\frac{n \pi}{4}+\frac{\pi}{16}$ or $x=n \pi-\frac{\pi}{4}$ where $n \in I$

So general solution is $x=\frac{n \pi}{4}+\frac{\pi}{16}$ or $x=n \pi-\frac{\pi}{4}$ where $n \in I$

(iii) Given: $\cos m x=\sin n x \Rightarrow \cos m x=\cos \left(\frac{\pi}{2}-n x\right)$

Formula used: $\cos \theta=\cos a \Rightarrow \theta=2 \mathrm{k} \pi \pm a, \mathrm{k} \in I$

By using the above formula, we have

$m x=2 k \pi+\left(\frac{\pi}{2}-n x\right)$ or $5 x=2 k \pi-\left(\frac{\pi}{2}-n x\right)$

$(m+n) x=2 k \pi+\frac{\pi}{2}$ or $(m-n) x=2 k \pi-\frac{\pi}{2}$

$x=\frac{2 k \pi}{(m+n)}+\frac{\pi}{2(m+n)}$ or $x=\frac{2 k \pi}{(m-n)}+\frac{\pi}{2(m-n)}$ where $k \in I$

$x=\frac{(4 k+1) \pi}{2(m+n)}$ or $x=\frac{(4 k-1) \pi}{2(m-n)}$ where $k \in I$

So the general solution is $x=\frac{(4 k+1) \pi}{2(\mathrm{~m}+\mathrm{n})}$ or $x=\frac{(4 k-1) \pi}{2(\mathrm{~m}-\mathrm{n})}$ where $k \in$ ।