Question

Solve the following equations:
(i) $\cot x+\tan x=2$ [NCERT EXEMPLAR]
(ii) $2{\sin }^{2}x=3\cos x,0\le x\le 2\mathrm{\pi }$ [NCERT EXEMPLAR]
(iii) $\sec x\cos 5x+1=0,0<x<\frac{\mathrm{\pi }}{2}$ [NCERT EXEMPLAR]
(iv) $5{\cos }^{2}x+7{\sin }^{2}x-6=0$ [NCERT EXEMPLAR]
(v) $\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$ [NCERT EXEMPLAR]
(vi) 4sinx cosx + 2 sin x + 2 cosx + 1 = 0
(vii) cosx + sin x = cos 2x + sin 2x
(viii) sin x tan x – 1 = tan x – sin x
(ix) 3tanx + cot x = 5 cosec x

Answer 1

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

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

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

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

(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. $\left(-1\le \cos x\le 1\right)$
$$\begin{gathered} \therefore \tan 2x=1=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathbf{Z} \\ \Rightarrow x=\frac{n\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{8},n\in \mathbf{Z} \end{gathered}$$

(vi)
$$\begin{gathered} 4\sin x\cos x+2\sin x+2\cos x+1=0 \\ \Rightarrow 2\sin x\left(2\cos x+1\right)+1\left(2\cos x+1\right)=0 \\ \Rightarrow \left(2\sin x+1\right)\left(2\cos x+1\right)=0 \\ \Rightarrow 2\sin x+1=0\mathrm{or}2\cos x+1=0 \\ \Rightarrow \sin x=-\frac{1}{2}\mathrm{or}\cos x=-\frac{1}{2} \\ \Rightarrow \sin x=\sin \frac{7\mathrm{\pi }}{6}\mathrm{or}\cos x=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6}\mathrm{or}x=2n\mathrm{\pi }\pm \frac{2\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}} \end{gathered}$$

(vii)
$$\begin{gathered} \cos x+\sin x=\cos 2x+\sin 2x \\ \Rightarrow \cos 2x-\cos x+\sin 2x-\sin x=0 \\ \Rightarrow -2\sin \frac{3x}{2}\sin \frac{x}{2}+2\cos \frac{3x}{2}\sin \frac{x}{2}=0 \\ \Rightarrow 2\sin \frac{x}{2}\left(\cos \frac{3x}{2}-\sin \frac{3x}{2}\right)=0 \\ \Rightarrow 2\sin \frac{x}{2}=0\mathrm{or}\cos \frac{3x}{2}-\sin \frac{3x}{2}=0 \\ \Rightarrow \sin \frac{x}{2}=0\mathrm{or}\cos \frac{3x}{2}=\sin \frac{3x}{2} \\ \Rightarrow \frac{x}{2}=n\mathrm{\pi }\mathrm{or}\tan \frac{3x}{2}=1 \\ \Rightarrow x=2n\mathrm{\pi }\mathrm{or}\tan \frac{3x}{2}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow x=2n\mathrm{\pi }\mathrm{or}\frac{3x}{2}=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }\mathrm{or}3\mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }\mathrm{or}\mathrm{x}=\frac{2\mathrm{n\pi }}{3}+\frac{\mathrm{\pi }}{6},n\in \mathrm{\mathbb{Z}} \end{gathered}$$

(viii)
$$\begin{gathered} \sin x\tan x-1=\tan x-\sin x \\ \Rightarrow \sin x\tan x-\tan x+\sin x-1=0 \\ \Rightarrow \tan x\left(\sin x-1\right)+1\left(\sin x-1\right)=0 \\ \Rightarrow \left(\tan x+1\right)\left(\sin x-1\right)=0 \\ \Rightarrow \left(\tan x+1\right)=0\mathrm{or}\left(\sin x-1\right)=0 \\ \Rightarrow \tan x=-1\mathrm{or}\sin x=1 \\ \Rightarrow \tan x=\tan \frac{3\mathrm{\pi }}{4}\mathrm{or}\sin x=\sin \frac{\mathrm{\pi }}{2} \\ \Rightarrow x=n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\mathrm{or}x=n\mathrm{\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{2},n\in \mathrm{\mathbb{Z}} \end{gathered}$$

(ix)
$$\begin{gathered} 3\tan x+\cot x=5\mathrm{cosec}x \\ \Rightarrow \frac{3\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{5}{\sin x} \\ \Rightarrow \frac{3{\sin }^{2}x+{\cos }^{2}x}{\cos x\sin x}=\frac{5}{\sin x} \\ \Rightarrow 3\left(1-{\cos }^{2}x\right)+{\cos }^{2}x=5\cos x \\ \Rightarrow 3-3{\cos }^{2}x+{\cos }^{2}x=5\cos x \\ \Rightarrow 2{\cos }^{2}x+5\cos x-3=0 \\ \Rightarrow 2{\cos }^{2}x+6\cos x-\cos x-3=0 \\ \Rightarrow 2\cos x\left(\cos x+3\right)-1\left(\cos x+3\right)=0 \\ \Rightarrow \left(2\cos x-1\right)\left(\cos x+3\right)=0 \\ \Rightarrow \left(2\cos x-1\right)=0\mathrm{or}\left(\cos x+3\right)=0 \\ \Rightarrow \cos x=\frac{1}{2}\mathrm{or}\cos x=-3 \\ \cos x=-3\mathrm{is}\mathrm{not}\mathrm{possible}\left(\because -1\le \cos x\le 1\right) \\ \Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},n\in \mathrm{\mathbb{Z}} \end{gathered}$$