Class 11 Maths Exercise 3.4 Solutions

$\tan x=\sqrt{3}$
it is known that $\tan \frac{\pi}{3}=\sqrt{3}$
and $\tan \left(\frac{4 \pi}{3}\right)=\tan \left(\pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3}$
Therefore, the principle solutions are $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$
Now, $\tan x=\tan \frac{\pi}{3}$
$=x=\mathrm{n} \pi+\frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
Therefore, the general solution is $x=\mathrm{n} \pi+\frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$

$\operatorname{Sec} x=2$
It is known that $\sec \frac{\pi}{3}=2$
and $\sec \frac{5 \pi}{3}=\sec \left(2 \pi-\frac{\pi}{3}\right)=\sec \frac{\pi}{3}=2$
Therefore, the principle solution are $x=\frac{\pi}{3}$ and $\frac{5 \pi}{3}$
Now, $\sec x=\sec \frac{\pi}{3}$
$\Rightarrow \cos x=\cos \frac{\pi}{3}$
$\Rightarrow x=2 \mathrm{n} \pi \pm \frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$
Therefore, the general solution is $x=2 \mathrm{n} \pi+\frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$

$\operatorname{Cot} x=-\sqrt{3}$
It is known that $\cot \frac{\pi}{6}=\sqrt{3}$
$\therefore \cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
and $\cot \left(2 \pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
= i.e., $\cot \frac{5 \pi}{6}=-\sqrt{3}$ and $\cot \frac{11 \pi}{6}=-\sqrt{3}$
Therefore, the principle solutions are $x=\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$
Now, $\cot x=\cot \frac{5 \pi}{6}$
$\Rightarrow \tan \mathrm{x}=\tan \frac{5 \pi}{6} \quad\left[\cot x=\frac{1}{\tan x}\right]$
$\Rightarrow x=\mathrm{n} \pi+\frac{5 \pi}{6}$, where $\mathrm{n} \in \mathrm{Z}$
Therefore, the general solution is $x=\mathrm{n} \pi+\frac{5 \pi}{6}$, where $\mathrm{n} \in \mathrm{Z}$

It is given that
$\operatorname{cosec} x=-2$
We know that
$\operatorname{sosec} \frac{\pi}{6}=2$
It can be written as
$\operatorname{cosec}\left(\pi+\frac{\pi}{6}\right)=-\operatorname{cosec} \frac{\pi}{6}=-2$
And
$\operatorname{cosec}\left(2 \pi-\frac{\pi}{6}\right)=-\operatorname{cosec} \frac{\pi}{6}=-2$
So we get
$\operatorname{cosec} \frac{7 \pi}{6}=-2$ and $\operatorname{cosec} \frac{11 \pi}{6}=-2$
Heace, the principal solutions are $x=\frac{ 7 \pi }{ 6 }$ and $\frac{ 11 \pi }{ 6 }$.
$\operatorname{cosec} x=\operatorname{cosec} \frac{7 \pi}{6}$

$\cos 4 x=\cos 2 x$
$=\cos 4 x-\cos 2 x=0$
$=-2 \sin \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)=0$
$=\left[\therefore \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A+B}{2}\right)\right]$
$=\sin 3 x \sin x=0$
$=\sin 3 x=0$ or $\sin x=0$
$\therefore 3 x=\mathrm{n} \pi$ or $x=\mathrm{n} \pi$, where $\mathrm{n} \in \mathrm{Z}$
$=x=\frac{n \pi}{3}$ or $x=\mathrm{n} \pi$, where $\mathrm{n} \in \mathrm{Z}$

$\cos 3 x+\cos x-\cos 2 x=0$
$\Rightarrow 2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)-\cos 2 x=0$
$\Rightarrow 2 \cos 2 x \cos x-\cos 2 x=0$
$\Rightarrow \cos 2 x(2 \cos x-1)=0$
$\Rightarrow \cos 2 x=0 \text { or } 2 \cos x-1=0$
$\Rightarrow \cos 2 x=0 \text { or } \cos x=\frac{1}{2}$
$\therefore 2 x=(2 \mathrm{n}+1) \frac{\pi}{2}$ or $\cos x=\cos \frac{\pi}{2}$, where $\mathrm{n} \in \mathrm{Z}$
$\Rightarrow x=(2 \mathrm{n}+1) \frac{\pi}{4}$ or $x=2 \mathrm{n} \pi \pm \frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$

$\sin ^{2} x+\cos x=0$
$\Rightarrow 2 \sin x \cos x+\cos x=0$
$\Rightarrow \cos x(2 \sin x+1)=0$
$\Rightarrow \cos x=0 \text { or } 2 \sin x+1=0$
Now, $\cos x=0 \Rightarrow \cos x=(2 \mathrm{n}+1) \frac{\pi}{2}$, where $\mathrm{n} \in \mathrm{Z}$
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}$
$\Rightarrow x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{7 \pi}{6}$, where $\mathrm{n} \in \mathrm{Z}$
Therefore, the general solution is $(2 \mathrm{n}+1) \frac{\pi}{2}$ or $\mathrm{n} \pi+(-1) ^\mathrm{n} \frac{7 \pi}{6}$, where $\mathrm{n} \in Z$

$\sec ^{2} 2 x=1-\tan 2 x$
$\Rightarrow 1+\tan ^{2} 2 x=1-\tan 2 x$
$\Rightarrow \tan ^{2} 2 x+\tan 2 x=0$
$\Rightarrow \tan 2 x(\tan 2 x+1)=0$
$\Rightarrow \tan 2 x=0 \text { or } \tan 2 x+1=0$
Now, $\tan 2 x=0$
$\Rightarrow \tan 2 x=\tan 0$
$\Rightarrow 2 x=\mathrm{n} \pi+0$, where $\mathrm{n} \in \mathrm{Z}$
$\Rightarrow x=\frac{n \pi}{2}$, where $\mathrm{n} \in \mathrm{Z}$
$\tan 2 x+1=0$
$\Rightarrow \tan 2 x=-1=-\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)=\tan \frac{3 \pi}{4}$
$\Rightarrow 2 x=\mathrm{n} \pi+\frac{3 \pi}{4}$, where $\mathrm{n} \in \mathrm{Z}$
$\Rightarrow x=\frac{n \pi}{2}+\frac{3 \pi}{8}$, where $\mathrm{n} \in \mathrm{Z}$
Therefore, the general solution is $\frac{n \pi}{2}$ or $\frac{n \pi}{2}+\frac{3 \pi}{8}, n \in \mathrm{z}$