Solve the following equations:


Solve the following equations:

(i) $\cot x+\tan x=2$                                                      [NCERT EXEMPLAR]

(ii) $2 \sin ^{2} x=3 \cos x, 0 \leq x \leq 2 \pi$                 [NCERT EXEMPLAR]

(iii) $\sec x \cos 5 x+1=0,0[NCERT EXEMPLAR]

(iv) $5 \cos ^{2} x+7 \sin ^{2} x-6=0$                                     [NCERT EXEMPLAR]

(v) $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$     [NCERT EXEMPLAR]

(vi) $4 \sin x \cos x+2 \sin x+2 \cos x+1=0$                           [NCERT EXEMPLAR]

(vii) $\cos x+\sin x=\cos 2 x+\sin 2 x$                                    [NCERT EXEMPLAR]

(viii) $\sin x \tan x-1=\tan x-\sin x$                                         [NCERT EXEMPLAR]

(ix) $3 \tan x+\cot x=5 \operatorname{cosec} x$                    [NCERT EXEMPLAR]


(i) $\cot x+\tan x=2$

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

$\Rightarrow \tan ^{2} x+1=2 \tan x$

$\Rightarrow \tan ^{2} x-2 \tan x+1=0$

$\Rightarrow(\tan x-1)^{2}=0$

$\Rightarrow \tan x=1=\tan \frac{\pi}{4}$

$\Rightarrow x=n \pi+\frac{\pi}{4}, n \in \mathbf{Z}$      $(\tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha, n \in \mathbf{Z})$

(ii) $2 \sin ^{2} x=3 \cos x$

$\Rightarrow 2\left(1-\cos ^{2} x\right)=3 \cos x$

$\Rightarrow 2 \cos ^{2} x+3 \cos x-2=0$

$\Rightarrow(2 \cos x-1)(\cos x+2)=0$

$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-2$

But, $\cos x=-2$ is not possible. $\quad(-1 \leq \cos x \leq 1)$

$\therefore \cos x=\frac{1}{2}=\cos \frac{\pi}{3}$

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

Putting $n=0$ and $n=1$, we get

$x=\frac{\pi}{3}, \frac{5 \pi}{3} \quad(0 \leq x \leq 2 \pi)$

(iii) $\sec x \cos 5 x+1=0$

$\Rightarrow \frac{\cos 5 x}{\cos x}+1=0$

$\Rightarrow \cos 5 x+\cos x=0$

$\Rightarrow 2 \cos 3 x \cos 2 x=0$

$\Rightarrow \cos 3 x=0$ or $\cos 2 x=0$

$\Rightarrow 3 x=(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}$ or $2 x=(2 m+1) \frac{\pi}{2}, m \in \mathbf{Z}$

$\Rightarrow \boldsymbol{x}=(2 n+1) \frac{\pi}{6}$ or $x=(2 m+1) \frac{\pi}{4}$

Putting n = 0 and = 0, we get

$x=\frac{\pi}{6}, \frac{\pi}{4} \quad\left(0

(iv) $5 \cos ^{2} x+7 \sin ^{2} x-6=0$

$\Rightarrow 5 \cos ^{2} x+7\left(1-\cos ^{2} x\right)-6=0$

$\Rightarrow-2 \cos ^{2} x+1=0$

$\Rightarrow \cos ^{2} x=\frac{1}{2}=\cos ^{2} \frac{\pi}{4}$

$\Rightarrow x=n \pi \pm \frac{\pi}{4}, n \in \mathbf{Z}$    $\left(\cos ^{2} \boldsymbol{x}=\cos ^{2} \alpha \Rightarrow x=n \pi \pm \alpha, n \in \mathbf{Z}\right)$

(v) $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$

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

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

$\Rightarrow(\sin 2 x-\cos 2 x)(2 \cos x-3)=0$

$\Rightarrow \sin 2 x-\cos 2 x=0$ or $2 \cos x-3=0$

$\Rightarrow \sin 2 x=\cos 2 x$ or $\cos x=\frac{3}{2}$

$\Rightarrow \tan 2 x=1$ or $\cos x=\frac{3}{2}$

But, $\cos x=\frac{3}{2}$ is not possible. $\quad(-1 \leq \cos x \leq 1)$

$\therefore \tan 2 x=1=\tan \frac{\pi}{4}$

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

$\Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{8}, n \in \mathbf{Z}$

(vi) $4 \sin x \cos x+2 \sin x+2 \cos x+1=0$

$\Rightarrow 2 \sin x(2 \cos x+1)+1(2 \cos x+1)=0$

$\Rightarrow(2 \sin x+1)(2 \cos x+1)=0$

$\Rightarrow 2 \sin x+1=0$ or $2 \cos x+1=0$

$\Rightarrow \sin x=-\frac{1}{2}$ or $\cos x=-\frac{1}{2}$

$\Rightarrow \sin x=\sin \frac{7 \pi}{6}$ or $\cos x=\frac{2 \pi}{3}$

$\Rightarrow x=n \pi+(-1)^{\mathrm{n}} \frac{7 \pi}{6}$ or $x=2 n \pi \pm \frac{2 \pi}{3}, n \in \mathbb{Z}$

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

$\Rightarrow \cos 2 x-\cos x+\sin 2 x-\sin x=0$

$\Rightarrow-2 \sin \frac{3 x}{2} \sin \frac{x}{2}+2 \cos \frac{3 x}{2} \sin \frac{x}{2}=0$

$\Rightarrow 2 \sin \frac{x}{2}\left(\cos \frac{3 x}{2}-\sin \frac{3 x}{2}\right)=0$

$\Rightarrow 2 \sin \frac{x}{2}=0$ or $\cos \frac{3 x}{2}-\sin \frac{3 x}{2}=0$

$\Rightarrow \sin \frac{x}{2}=0$ or $\cos \frac{3 x}{2}=\sin \frac{3 x}{2}$

$\Rightarrow \frac{x}{2}=n \pi$ or $\tan \frac{3 x}{2}=1$

$\Rightarrow x=2 n \pi$ or $\tan \frac{3 x}{2}=\tan \frac{\pi}{4}$

$\Rightarrow x=2 n \pi$ or $\frac{3 x}{2}=n \pi+\frac{\pi}{4}$

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

$\Rightarrow \mathrm{x}=2 \mathrm{n} \pi$ or $\mathrm{x}=\frac{2 \mathrm{n} \pi}{3}+\frac{\pi}{6}, n \in \mathbb{Z}$

(viii) $\sin x \tan x-1=\tan x-\sin x$

$\Rightarrow \sin x \tan x-\tan x+\sin x-1=0$

$\Rightarrow \tan x(\sin x-1)+1(\sin x-1)=0$

$\Rightarrow(\tan x+1)(\sin x-1)=0$

$\Rightarrow(\tan x+1)=0$ or $(\sin x-1)=0$

$\Rightarrow \tan x=-1$ or $\sin x=1$

$\Rightarrow \tan x=\tan \frac{3 \pi}{4}$ or $\sin x=\sin \frac{\pi}{2}$

$\Rightarrow x=n \pi+\frac{3 \pi}{4}$ or $x=n \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}, n \in \mathbb{Z}$

(ix) $3 \tan x+\cot x=5 \operatorname{cosec} x$

$\Rightarrow \frac{3 \sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{5}{\sin x}$

$\Rightarrow \frac{3 \sin ^{2} x+\cos ^{2} x}{\cos x \sin x}=\frac{5}{\sin x}$

$\Rightarrow 3\left(1-\cos ^{2} x\right)+\cos ^{2} x=5 \cos x$

$\Rightarrow 3-3 \cos ^{2} x+\cos ^{2} x=5 \cos x$

$\Rightarrow 2 \cos ^{2} x+5 \cos x-3=0$

$\Rightarrow 2 \cos ^{2} x+6 \cos x-\cos x-3=0$

$\Rightarrow 2 \cos x(\cos x+3)-1(\cos x+3)=0$

$\Rightarrow(2 \cos x-1)(\cos x+3)=0$

$\Rightarrow(2 \cos x-1)=0$ or $(\cos x+3)=0$

$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-3$

$\cos x=-3$ is not possible $\quad(\because-1 \leq \cos x \leq 1)$

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

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now