# Find the general solutions of the following equations:

Question:

Find the general solutions of the following equations:

(i) $\sin 2 x=\frac{\sqrt{3}}{2}$

(ii) $\cos 3 x=\frac{1}{2}$

(iii) $\sin 9 x=\sin \mathrm{x}$

(iv) $\sin 2 x=\cos 3 x$

(v) $\tan x+\cot 2 x=0$

(vi) $\tan 3 x=\cot x$

(vii) $\tan 2 x \tan x=1$

(viii) $\tan m x+\cot n x=0$

(ix) $\tan p x=\cot q x$

(x) $\sin 2 x+\cos x=0$

(xi) $\sin x=\tan x$

(xii) $\sin 3 x+\cos 2 x=0$

Solution:

We have:

(i) $\sin 2 x=\frac{\sqrt{3}}{2}$

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

$\Rightarrow 2 x=n \pi+(-1)^{\mathrm{n}} \frac{\pi}{3} n \in Z$

$\Rightarrow x=\frac{n \pi}{2}+(-1)^{n} \frac{\pi}{6}, n \in Z$

(ii) $\cos 3 x=\frac{1}{2}$

$\Rightarrow \cos 3 x=\cos \frac{\pi}{3}$

$\Rightarrow 3 x=2 n \pi \pm \frac{\pi}{3} n \in Z$

$\Rightarrow x=\frac{2 n \pi}{3} \pm \frac{\pi}{9}, n \in Z$

(iii) $\sin 9 x=\sin x$

$\Rightarrow \sin 9 x-\sin x=0$

$\Rightarrow 2 \sin \left(\frac{9 x-x}{2}\right) \cos \left(\frac{9 x+x}{2}\right)=0$

$\Rightarrow \sin \frac{8 x}{2}=0$ or $\cos \frac{10 x}{2}=0$

$\Rightarrow \sin 4 x=0$ or $\cos 5 x=0$

$\Rightarrow 4 x=n \pi, n \in Z$ or $5 x=(2 n+1) \frac{\pi}{2}, n \in Z$

$\Rightarrow x=\frac{n \pi}{4}, n \in Z$ or $x=(2 n+1) \frac{\pi}{10}, n \in Z$

(iv) $\sin 2 x=\cos 3 x$

$\cos 3 x=\sin 2 x$

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

$\Rightarrow 3 x=2 n \pi \pm\left(\frac{\pi}{2}-2 x\right), n \in Z$

On taking positive sign, we have:

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

$\Rightarrow 5 x=2 n \pi+\frac{\pi}{2}$

$\Rightarrow x=\frac{2 \mathrm{n} \pi}{5}+\frac{\pi}{10}$

$\Rightarrow x=(4 n+1) \frac{\pi}{10}, n \in Z$

Now, on taking negative sign, we have:

$3 x=2 n \pi-\frac{\pi}{2}+2 x, n \in Z$

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

$\Rightarrow x=(4 n-1) \frac{\pi}{2}, n \in Z$

(v) $\tan x+\cot 2 x=0$

$\Rightarrow \tan x=-\cot 2 x$

$\Rightarrow \tan x=\tan \left(\frac{\pi}{2}+2 x\right)$

$\Rightarrow x=n \pi+\left(\frac{\pi}{2}+2 x\right), n \in Z$

$\Rightarrow-x=n \pi+\frac{\pi}{2}, n \in Z$

$\Rightarrow x=-n \pi-\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

$\Rightarrow \mathrm{x}=\mathrm{m} \pi-\frac{\pi}{2}, m=-\mathrm{n} \in \mathrm{Z}$

(vi) $\tan 3 x=\cot x$

$\Rightarrow \tan 3 x=\tan \left(\frac{\pi}{2}-x\right)$

$\Rightarrow 3 x=n \pi+\left(\frac{\pi}{2}-x\right), n \in Z$

$\Rightarrow 4 x=n \pi+\frac{\pi}{2}, n \in Z$

$\Rightarrow x=\frac{n \pi}{4}+\frac{\pi}{8}, n \in Z$

(vii) $\tan 2 x \tan x=1$

$\Rightarrow \tan 2 x=\frac{1}{\tan x}$

$\Rightarrow \tan 2 x=\cot x$

$\Rightarrow \tan 2 x=\tan \left(\frac{\pi}{2}-x\right)$

$\Rightarrow 2 x=n \pi+\left(\frac{\pi}{2}-x\right), n \in Z$

$\Rightarrow 3 x=n \pi+\frac{\pi}{2}, n \in Z$

$\Rightarrow x=\frac{n \pi}{3}+\frac{\pi}{6}, n \in Z$

(viii) $\tan m x+\cot n x=0$

$\Rightarrow \tan m x=-\cot n x$

$\Rightarrow \tan m x=\tan \left(\frac{\pi}{2}+n x\right)$

$\Rightarrow m x=r \pi+\left(\frac{\pi}{2}+n x\right), r \in Z$

$\Rightarrow(m-n) x=r \pi+\frac{\pi}{2}, r \in Z$

$\Rightarrow x=\left(\frac{2 r+1}{m-n}\right) \frac{\pi}{2}, r \in Z$

(ix) $\tan p x=\cot q x$

$\Rightarrow \tan p x=\tan \left(\frac{\pi}{2}-q x\right)$

$\Rightarrow p x=n \pi+\left(\frac{\pi}{2}-q x\right), n \in Z$

$\Rightarrow(p+q) x=n \pi+\frac{\pi}{2}, n \in Z$

$\Rightarrow x=\left(\frac{2 n+1}{p+q}\right) \frac{\pi}{2}, n \in Z$

(x) $\sin 2 x+\cos x=0$

$\Rightarrow \cos x=-\sin 2 x$

$\Rightarrow \cos x=\cos \left(\frac{\pi}{2}+2 x\right)$

$\Rightarrow x=2 n \pi \pm\left(\frac{\pi}{2}+2 x\right), n \in Z$

On taking positive sign, we have:

$x=2 n \pi+\frac{\pi}{2}+2 x$

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

$\Rightarrow x=2 m \pi-\frac{\pi}{2}, m=-n \in Z$

$\Rightarrow x=\frac{(4 m-1) \pi}{2}, m \in Z$

On taking negative sign, we have:

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

$\Rightarrow 3 x=2 n \pi-\frac{\pi}{2}$

$\Rightarrow x=\frac{(4 n-1) \pi}{6}, n \in \mathrm{Z}$

(xi) $\sin x=\tan x$

$\Rightarrow \sin x-\tan x=0$

$\Rightarrow \sin x-\frac{\sin x}{\cos x}=0$

$\Rightarrow \sin x\left(1-\frac{1}{\cos x}\right)=0$

$\Rightarrow \sin x(\cos x-1)=0$

$\Rightarrow \sin x=0$ or $\cos x-1=0$

Now,

$\sin x=0 \Rightarrow x=n \pi, n \in Z$

$\cos x-1=0$

$\Rightarrow \cos x=1$

$\Rightarrow \cos x=\cos 0$

$\Rightarrow x=2 m \pi, m \in Z$

(xii) $\sin 3 x+\cos 2 x=0$

$\Rightarrow \cos 2 x=-\sin 3 x$

$\Rightarrow \cos 2 x=\cos \left(\frac{\pi}{2}+3 x\right)$

$\Rightarrow 2 x=2 n \pi \pm\left(\frac{\pi}{2}+3 x\right), \mathrm{n} \in \mathrm{Z}$

On taking positive sign, we have:

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

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

$\Rightarrow x=2 m \pi-\frac{\pi}{2}, m=-n \in Z$

$\Rightarrow x=\frac{(4 m-1) \pi}{2}, m \in Z$

On taking negative sign, we have:

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

$\Rightarrow 5 x=2 n \pi-\frac{\pi}{2}$

$\Rightarrow x=\frac{(4 n-1) \pi}{10}, n \in Z$