Question

Solve the following equations:

(i) $\cot \theta +\tan \theta =2$ [NCERT EXEMPLAR]
(ii) $2{\sin }^2\theta =3\cos \theta ,0\le \theta \le 2\mathrm{\pi }$ [NCERT EXEMPLAR]
(iii) $\sec \theta \cos 5\theta +1=0,0<\theta <\frac{\mathrm{\pi }}{2}$ [NCERT EXEMPLAR]
(iv) $5{\cos }^2\theta +7{\sin }^2\theta -6=0$ [NCERT EXEMPLAR]
(v) $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ [NCERT EXEMPLAR]

Answer 1

(i)
$$\begin{gathered} \cot \theta +\tan \theta =2 \\ \Rightarrow \frac{1}{\tan \theta }+\tan \theta =2 \\ \Rightarrow {\tan }^2\theta +1=2\tan \theta \\ \Rightarrow {\tan }^2\theta -2\tan \theta +1=0 \\ \Rightarrow {\left(\tan \theta -1\right)}^2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =1=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta =n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathbf{Z}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(\tan \theta =\tan \alpha \Rightarrow \theta =n\mathrm{\pi }+\alpha ,n\in \mathbf{Z}\right) \end{gathered}$$

(ii)
$$\begin{gathered} 2{\sin }^2\theta =3\cos \theta \\ \Rightarrow 2\left(1-{\cos }^2\theta \right)=3\cos \theta \\ \Rightarrow 2{\cos }^2\theta +3\cos \theta -2=0 \\ \Rightarrow \left(2\cos \theta -1\right)\left(\cos \theta +2\right)=0 \end{gathered}$$
$\Rightarrow \cos \theta =\frac{1}{2}\mathrm{or}\cos \theta =-2$
But, $\cos \theta =-2$ is not possible. $\left(-1\le \cos \theta \le 1\right)$
$$\begin{gathered} \therefore \cos \theta =\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathbf{Z} \end{gathered}$$
Putting n = 0 and n = 1, we get
$\theta =\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\left(0\le \theta \le 2\mathrm{\pi }\right)$

(iii)
$$\begin{gathered} \sec \theta \cos 5\theta +1=0 \\ \Rightarrow \frac{\cos 5\theta }{\cos \theta }+1=0 \\ \Rightarrow \cos 5\theta +\cos \theta =0 \\ \Rightarrow 2\cos 3\theta \cos 2\theta =0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \cos 3\theta =0\mathrm{or}\cos 2\theta =0 \\ \Rightarrow 3\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in \mathbf{Z}\mathrm{or}2\theta =\left(2m+1\right)\frac{\mathrm{\pi }}{2},m\in \mathbf{Z} \\ \mathbf{\Rightarrow }\theta =\left(2n+1\right)\frac{\mathrm{\pi }}{6}\mathrm{or}\theta =\left(2m+1\right)\frac{\mathrm{\pi }}{4} \end{gathered}$$
Putting n = 0 and m = 0, we get
$\theta =\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{4}\left(0<\theta <\frac{\mathrm{\pi }}{2}\right)$

(iv)
$$\begin{gathered} 5{\cos }^2\theta +7{\sin }^2\theta -6=0 \\ \Rightarrow 5{\cos }^2\theta +7\left(1-{\cos }^2\theta \right)-6=0 \\ \Rightarrow -2{\cos }^2\theta +1=0 \\ \Rightarrow {\cos }^2\theta =\frac{1}{2}={\cos }^2\frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta =n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},n\in \mathbf{Z}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left({\cos }^{\mathbf{2}}\theta ={\cos }^2\alpha \Rightarrow \theta =n\mathrm{\pi }\pm \alpha ,n\in \mathbf{Z}\mathbf{}\right) \end{gathered}$$

(v)
$$\begin{gathered} \sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x \\ \Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x \\ \Rightarrow \sin 2x\left(2\cos x-3\right)=\cos 2x\left(2\cos x-3\right) \\ \Rightarrow \left(\sin 2x-\cos 2x\right)\left(2\cos x-3\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \sin 2x-\cos 2x=0\mathrm{or}2\cos x-3=0 \\ \Rightarrow \sin 2x=\cos 2x\mathrm{or}\cos x=\frac{3}{2} \\ \Rightarrow \tan 2x=1\mathrm{or}\cos x=\frac{3}{2} \end{gathered}$$
But, $\cos x=\frac{3}{2}$ is not possible. $\left(-1\le \cos x\le 1\right)$
$$\begin{gathered} \therefore \tan 2x=1=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathbf{Z} \\ \Rightarrow x=\frac{n\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{8},n\in \mathbf{Z} \end{gathered}$$