Question

Find the general solutions of the following equations:
(i) $\sin 2x=\frac{\sqrt{3}}{2}$
(ii) $\cos 3x=\frac{1}{2}$
(iii) $\sin 9x=\sin \mathrm{x}$
(iv) $\sin 2x=\cos 3x$
(v) $\tan x+\cot 2x=0$
(vi) $\tan 3x=\cot x$
(vii) $\tan 2x\tan x=1$
(viii) $\tan mx+\cot nx=0$
(ix) $\tan px=\cot qx$
(x) $\sin 2x+\cos x=0$
(xi) $\sin x=\tan x$
(xii) $\sin 3x+\cos 2x=0$

Answer 1

We have:

(i) $\sin 2x=\frac{\sqrt{3}}{2}$
⇒ $\sin 2x=\sin \frac{\mathrm{\pi }}{3}$
⇒ $2x=n\mathrm{\pi }+(-1{)}^{\mathrm{n}}\frac{\mathrm{\pi }}{3}$ $n\in Z$
⇒ $x=\frac{n\mathrm{\pi }}{2}+(-1{)}^n\frac{\mathrm{\pi }}{6}$, $n\in Z$

(ii) $\cos 3x=\frac{1}{2}$
⇒ $\cos 3x=\cos \frac{\mathrm{\pi }}{3}$
⇒ $3x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$ $n\in Z$
⇒ $x=\frac{2n\mathrm{\pi }}{3}\pm \frac{\mathrm{\pi }}{9}$, $n\in Z$

(iii) $\sin 9x=\sin x$
⇒ $\sin 9x-\sin x=0$
⇒ $2\sin \left(\frac{9x-x}{2}\right)\cos \left(\frac{9x+x}{2}\right)=0$
⇒ $\sin \frac{8x}{2}=0$or $\cos \frac{10x}{2}=0$
⇒ $\sin 4x=0$ or $\cos 5x=0$
⇒ $4x=n\mathrm{\pi }$, $n\in Z$ or $5x=(2n+1)\frac{\mathrm{\pi }}{2}$, $n\in Z$
⇒ $x=\frac{n\mathrm{\pi }}{4}$, $n\in Z$ or $x=(2n+1)\frac{\mathrm{\pi }}{10}$, $n\in Z$

(iv) $\sin 2x=\cos 3x$
$\cos 3x=\sin 2x$
⇒ $\cos 3x=\cos \left(\frac{\mathrm{\pi }}{2}-2x\right)$
⇒$3x=2n\mathrm{\pi }\pm \left(\frac{\mathrm{\pi }}{2}-2x\right),n\in Z$
On taking positive sign, we have:
$3x=2n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{2}-2x\right)$
⇒ $5x=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}$
⇒ $x=\frac{2\mathrm{n}\mathrm{\pi }}{5}+\frac{\mathrm{\pi }}{10}$
⇒$x=(4n+1)\frac{\mathrm{\pi }}{10}$, $n\in Z$
Now, on taking negative sign, we have:
$3x=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}+2x,n\in Z$
⇒ $x=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}$
⇒ $x=(4n-1)\frac{\mathrm{\pi }}{2},n\in Z$

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

$$\begin{gathered} \Rightarrow \tan x=-\cot 2x \\ \Rightarrow \tan x=\tan \left(\frac{\mathrm{\pi }}{2}+2x\right) \\ \Rightarrow x=n\pi +\left(\frac{\mathrm{\pi }}{2}+2x\right),n\in Z \\ \Rightarrow -x=n\pi +\frac{\mathrm{\pi }}{2},n\in Z \\ \Rightarrow x=-n\mathrm{\pi }-\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z} \\ \Rightarrow \mathrm{x}=\mathrm{m\pi }-\frac{\mathrm{\pi }}{2},m=-\mathrm{n}\in \mathrm{Z} \end{gathered}$$

(vi) $\tan 3x=\cot x$
$$\begin{gathered} \Rightarrow \tan 3x=\tan \left(\frac{\mathrm{\pi }}{2}-x\right) \\ \Rightarrow 3x=n\pi +\left(\frac{\mathrm{\pi }}{2}-x\right),n\in Z \\ \Rightarrow 4x=n\pi +\frac{\mathrm{\pi }}{2},n\in Z \\ \Rightarrow x=\frac{n\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{8},n\in Z \end{gathered}$$

(vii) $\tan 2x\tan x=1$
$$\begin{gathered} \Rightarrow \tan 2x=\frac{1}{\tan x} \\ \Rightarrow \tan 2x=\cot x \\ \Rightarrow \tan 2x=\tan \left(\frac{\mathrm{\pi }}{2}-x\right) \\ \Rightarrow 2x=n\pi +\left(\frac{\mathrm{\pi }}{2}-x\right),n\in Z \\ \Rightarrow 3x=n\pi +\frac{\mathrm{\pi }}{2},n\in Z \\ \Rightarrow x=\frac{n\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6},n\in Z \end{gathered}$$

(viii) $\tan mx+\cot nx=0$

$$\begin{gathered} \Rightarrow \tan mx=-\cot nx \\ \Rightarrow \tan mx=\tan \left(\frac{\mathrm{\pi }}{2}+nx\right) \\ \Rightarrow mx=r\pi +\left(\frac{\mathrm{\pi }}{2}+nx\right),r\in Z \\ \Rightarrow (m-n)x=r\pi +\frac{\mathrm{\pi }}{2},r\in Z \\ \Rightarrow x=\left(\frac{2r+1}{m-n}\right)\frac{\mathrm{\pi }}{2},r\in Z \end{gathered}$$

(ix) $\tan px=\cot qx$
$$\begin{gathered} \Rightarrow \tan px=\tan \left(\frac{\mathrm{\pi }}{2}-qx\right) \\ \Rightarrow px=n\pi +\left(\frac{\mathrm{\pi }}{2}-qx\right),n\in Z \\ \Rightarrow (p+q)x=n\pi +\frac{\mathrm{\pi }}{2},n\in Z \\ \Rightarrow x=\left(\frac{2n+1}{p+q}\right)\frac{\mathrm{\pi }}{2},n\in Z \end{gathered}$$

(x) $\sin 2x+\cos x=0$
$$\begin{gathered} \Rightarrow \cos x=-\sin 2x \\ \Rightarrow \cos x=\cos \left(\frac{\mathrm{\pi }}{2}+2x\right) \\ \Rightarrow x=2n\pi \pm \left(\frac{\mathrm{\pi }}{2}+2x\right),n\in Z \end{gathered}$$
On taking positive sign, we have:
$$\begin{gathered} x=2n\pi +\frac{\mathrm{\pi }}{2}+2x \\ \Rightarrow -x=2n\pi +\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=2m\pi -\frac{\mathrm{\pi }}{2},m=-n\in Z \\ \Rightarrow x=\frac{(4m-1)\mathrm{\pi }}{2},m\in Z \end{gathered}$$
On taking negative sign, we have:
$$\begin{gathered} x=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}-2x \\ \Rightarrow 3x=2n\pi -\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=\frac{(4n-1)\mathrm{\pi }}{6},n\in \mathrm{Z} \end{gathered}$$

(xi) $\sin x=\tan x$
$$\begin{gathered} \Rightarrow \sin x-\tan x=0 \\ \Rightarrow \sin x-\frac{\sin x}{\cos x}=0 \\ \Rightarrow \sin x\left(1-\frac{1}{\cos x}\right)=0 \\ \Rightarrow \sin x(\cos x-1)=0 \end{gathered}$$

$\Rightarrow \sin x=0$ or $\cos x-1=0$
Now,
$\sin x=0\Rightarrow x=n\pi ,n\in Z$

$$\begin{gathered} \cos x-1=0 \\ \Rightarrow \cos x=1 \\ \Rightarrow \cos x=\cos 0 \\ \Rightarrow x=2m\pi ,m\in Z \end{gathered}$$

(xii) $\sin 3x+\cos 2x=0$
$$\begin{gathered} \Rightarrow \cos 2x=-\sin 3x \\ \Rightarrow \cos 2x=\cos \left(\frac{\mathrm{\pi }}{2}+3x\right) \\ \Rightarrow 2x=2n\mathrm{\pi }\pm \left(\frac{\mathrm{\pi }}{2}+3x\right),\mathrm{n}\in \mathrm{Z} \end{gathered}$$
On taking positive sign, we have:
$$\begin{gathered} 2x=2n\pi +\frac{\mathrm{\pi }}{2}+3x \\ \Rightarrow -x=2n\pi +\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=2m\pi -\frac{\mathrm{\pi }}{2},m=-n\in Z \\ \Rightarrow x=\frac{(4m-1)\mathrm{\pi }}{2},m\in Z \end{gathered}$$
On taking negative sign, we have:
$$\begin{gathered} 2x=2n\pi -\frac{\mathrm{\pi }}{2}-3x \\ \Rightarrow 5x=2n\pi -\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=\frac{(4n-1)\mathrm{\pi }}{10},n\in Z \end{gathered}$$