Question

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

Answer 1

We have:

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

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

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


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

Now, on taking negative sign, we have:
$3\theta =2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}+2\mathrm{\theta },n\in Z$
⇒ $\theta =2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}$

⇒ $\theta =(4n-1)\frac{\mathrm{\pi }}{2},n\in Z$


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

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

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

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

(viii) $\tan m\theta +cotn\theta =0$

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

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

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

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

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

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

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