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Looking for Class 11 Maths Exercise 3.4 solutions? You’re in the right place! This section provides detailed and accurate solutions to all the questions from Exercise 3.4 Class 11 Maths, part of Chapter 3 – Trigonometric Functions. These step-by-step explanations are based on the latest NCERT solutions for Class 11 Maths Chapter 3 and are ideal for students looking to understand the properties and graphs of trigonometric functions. Whether you’re searching for class 11 ch 3 exercise 3.4 solutions or practicing from the NCERT Exemplar Class 11 Maths, these resources will help you grasp key concepts with clarity. Master trigonometric functions Class 11 with these well-structured and exam-ready solutions. Download or view now to boost your preparation!
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Exercise 3.4
1. Find the principal and general solutions of the following equations: \( \tan x=\sqrt{3} \)
Answer
\(\tan x=\sqrt{3}\)
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 \( \mathrm{n} \in \mathrm{Z} \)
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2. Find the principal and general solutions of the following equations: \( \sec x=2 \)
Answer
\( \operatorname{Sec} x=2 \)
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 \( \mathrm{n} \in \mathrm{Z} \)
3. Find the principal and general solutions of the following equations: \( \cot x=-\sqrt{3} \)
Answer
\( \operatorname{Cot} x=-\sqrt{3} \)
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 \( \mathrm{n} \in \mathrm{Z} \)
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4. Find the general solution for each of the following equations: \( \operatorname{cosec} x=-2 \)
Answer
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 \( \frac{ 11 \pi }{ 6 } \).
\(\operatorname{cosec} x=\operatorname{cosec} \frac{7 \pi}{6}\)
5. Find the general solution for each of the following equations:
\(\cos 4 x=\cos 2 x\)
Answer
\(\cos 4 x=\cos 2 x\)
\(=\cos 4 x-\cos 2 x=0\)
\(=-2 \sin \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)=0\)
\(=\left[\therefore \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A+B}{2}\right)\right]\)
\(=\sin 3 x \sin x=0\)
\( =\sin 3 x=0 \) 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 \( \mathrm{n} \in \mathrm{Z} \)
6. Find the general solution for each of the following equations:
\(\cos 3 x+\cos x-\cos 2 x=0\)
Answer
\(\cos 3 x+\cos x-\cos 2 x=0\)
\(\Rightarrow 2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)-\cos 2 x=0\)
\(\Rightarrow 2 \cos 2 x \cos x-\cos 2 x=0\)
\(\Rightarrow \cos 2 x(2 \cos x-1)=0\)
\(\Rightarrow \cos 2 x=0 \text { or } 2 \cos x-1=0\)
\(\Rightarrow \cos 2 x=0 \text { or } \cos x=\frac{1}{2}\)
\( \therefore 2 x=(2 \mathrm{n}+1) \frac{\pi}{2} \) 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 \( \mathrm{n} \in \mathrm{Z} \)
7. Find the general solution for each of the following equations: \(\sin ^{2} x+\cos x=0\)
Answer
\(\sin ^{2} x+\cos x=0\)
\(\Rightarrow 2 \sin x \cos x+\cos x=0\)
\(\Rightarrow \cos x(2 \sin x+1)=0\)
\(\Rightarrow \cos x=0 \text { or } 2 \sin x+1=0\)
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 \( \mathrm{n} \in Z\)
8. Find the general solution for each of the following equations: \(\sec ^{2} 2 x=1-\tan 2 x\)
Answer
\(\sec ^{2} 2 x=1-\tan 2 x\)
\(\Rightarrow 1+\tan ^{2} 2 x=1-\tan 2 x\)
\(\Rightarrow \tan ^{2} 2 x+\tan 2 x=0\)
\(\Rightarrow \tan 2 x(\tan 2 x+1)=0\)
\(\Rightarrow \tan 2 x=0 \text { or } \tan 2 x+1=0\)
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 \( \frac{n \pi}{2}+\frac{3 \pi}{8}, n \in \mathrm{z} \)
