NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.4

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Chapter 3 Ex.3.4 Question 1

Find the principal and general solutions of the following equation:

\(\tan x = \sqrt 3 \)

Solution

It is given that \(\tan x = \sqrt 3 \)

We know that \(\tan \frac{\pi }{3} = \sqrt 3 \) and

\[\begin{align}\tan \frac{{4\pi }}{3} &= \tan \left( {\pi + \frac{\pi }{3}} \right)\\&= \tan \frac{\pi }{3}\\&= \sqrt 3 \end{align}\]

Hence the principal solutions are \(x = \frac{\pi }{3}\) and \(\frac{{4\pi }}{3}\).

Now,

\(\tan x = \tan \frac{\pi }{3}\)

Therefore, \(x = n\pi + \frac{\pi }{3}\), where \(n \in Z\).

Hence the general solution is \(x = n\pi + \frac{\pi }{3}\), where \(n \in Z\).

Chapter 3 Ex.3.4 Question 2

Question 2:

Find the principal and general solutions of the following equation:

\(\sec x = 2\)

Solution

It is given that \(\sec x = 2\)

We know that \(\sec \frac{\pi }{3} = 2\) and

\[\begin{align}\sec \frac{{5\pi }}{3} &= \sec \left( {2\pi - \frac{\pi }{3}} \right)\\&= \sec \frac{\pi }{3}\\&= \sqrt 3 \end{align}\]

Hence the principal solutions are \(x = \frac{\pi }{3}\) and \(\frac{{5\pi }}{3}\).

Now,

\(\begin{array}{l}\sec x = \sec \frac{\pi }{3}\\\cos x = \cos \frac{\pi }{3}\end{array}\)

Therefore, \(x = 2n\pi \pm \frac{\pi }{3}\), where \(n \in Z\).

Hence the general solution is \(x = 2n\pi \pm \frac{\pi }{3}\), where \(n \in Z\).

Chapter 3 Ex.3.4 Question 3

Find the principal and general solutions of the following equation:

\(\cot x = - \sqrt 3 \)

Solution

It is given that \(\cot x = - \sqrt 3 \)

We know that \(\cot \frac{\pi }{6} = \sqrt 3 \)

Therefore,

\[\begin{align}\cot \left( {\pi - \frac{\pi }{6}} \right) &= - \cot \frac{\pi }{6}\\\cot \left( {\frac{{5\pi }}{6}} \right) &= - \sqrt 3 \end{align}\]

and

\[\begin{align}\cot \left( {2\pi - \frac{\pi }{6}} \right) &= - \cot \frac{\pi }{6}\\\cot \left( {\frac{{11\pi }}{6}} \right)& = - \sqrt 3 \end{align}\]

Hence the principal solutions are \(x = \frac{{5\pi }}{6}\) and \(\frac{{11\pi }}{6}\).

Now,

\[\begin{align}\cot x &= \cot \left( {\frac{{5\pi }}{6}} \right)\\\tan x &= \tan \left( {\frac{{5\pi }}{6}} \right)\end{align}\]

Therefore, \(x = n\pi + \frac{{5\pi }}{6}\), where \(n \in Z\).

Hence the general solution is \(x = n\pi + \frac{{5\pi }}{6}\), where \(n \in Z\).

Chapter 3 Ex.3.4 Question 4

Find the principal and general solutions of the following equation:

\(\text{cosec}x = - 2\)

Solution

It is given that \(\text{cosec}x = - 2\)

We know that \(\text{cosec} \frac{\pi }{6} = 2\) and

Therefore,

\[\begin{align}{\text{cosec}}\ \left( {\pi + \frac{\pi }{6}} \right) &= - {\text{cosec}}\ \frac{\pi }{6}\\{\text{cosec}}\ \left( {\frac{{7\pi }}{6}} \right) &= - 2\end{align}\]

and

\[\begin{align}{\text{cosec}}\ \left( {2\pi - \frac{\pi }{6}} \right) &= - {\text{cosec}} \frac{\pi }{6}\\{\rm{co}}\sec \left( {\frac{{11\pi }}{6}} \right) &= - 2\end{align}\]

Hence the principal solutions are \(x = \frac{{7\pi }}{6}\) and \(\frac{{11\pi }}{6}\).

Now,

\[\begin{align}{\text{cosec}} x &= {\rm{co}}\sec \frac{{7\pi }}{6}\\\sin x& = \sin \frac{{7\pi }}{6}\end{align}\]

Therefore, \(x = n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6}\), where \(n \in Z\).

Hence the general solution is \(x = n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6}\), where \(n\in Z\).

Chapter 3 Ex.3.4 Question 5

Find the principal and general solutions of the following equation:

\(\cos 4x=\cos 2x\)

Solution

\[\begin{align} & \cos 4x-\cos 2x=0 \\ & -2\sin \left( \frac{4x+2x}{2} \right)\sin \left( \frac{4x-2x}{2} \right)=0\\&\qquad \left[ \because \cos A-\cos B=-2\sin \left( \frac{A+B}{2} \right)\sin \left( \frac{A-B}{2} \right) \right] \\ & \sin 3x\sin x=0 \\ \end{align}\]

\(\sin 3x = 0\) or \(\sin x = 0\)

\(3x = n\pi \) or \(x = n\pi \) where \(n \in Z\)

\(x = \frac{{n\pi }}{3}\) or \(x = n\pi \)

Therefore, \(x = \frac{{n\pi }}{3}\) or \(n\pi \), where \(n \in Z\)

Chapter 3 Ex.3.4 Question 6

Find the principal and general solutions of the following equation:

\(\cos 3x + \cos x - \cos 2x = 0\)

Solution

\[\begin{align} & \cos 3x+\cos x-\cos 2x=0 \\ & 2\cos \left( \frac{3x+x}{2} \right)\cos \left( \frac{3x-x}{2} \right)-\cos 2x=0\\& \quad \qquad \left[ \because \cos A+\cos B=2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ & 2\cos 2x\cos x-\cos 2x=0 \\ & \cos 2x\left( 2\cos x-1 \right)=0 \\ \end{align}\]

\(\cos 2x = 0\) or \(2\cos x - 1 = 0\)

\(2x = \left( {2n + 1} \right)\frac{\pi }{2}\) or \(\cos x = \frac{1}{2}\)

\(x = \left( {2n + 1} \right)\frac{\pi }{4}\) or \(x = 2n\pi \pm \frac{\pi }{3}\)

Therefore, \(x = \left( {2n + 1} \right)\frac{\pi }{4}\) or \(\left( {2n\pi \pm \frac{\pi }{3}} \right)\), where \(n \in Z\)

Chapter 3 Ex.3.4 Question 7

Find the principal and general solutions of the following equation:

\(\sin 2x + \cos x = 0\)

Solution

\[\begin{align} & \sin 2x+\cos x=0 \\ & 2\sin x\cos x+\cos x=0\\& \quad \qquad \left[ \because \sin 2x=2\sin x\cos x \right] \\ & \cos x\left( 2\sin x+1 \right)=0 \\ \end{align}\]

Now,

\(\cos x = 0\)     or     \(2\sin x + 1 = 0\)

Therefore,

\[\begin{align}\cos x &= 0\\x &= \left( {2n + 1} \right)\frac{\pi }{2} \quad n \in Z\end{align}\]

Or,

\[\begin{align}2\sin x + 1 &= 0\\\sin x &= - \frac{1}{2}\\&= - \sin \frac{\pi }{6}\\&= \sin \left( {\pi + \frac{\pi }{6}} \right)\end{align}\]

\[\begin{align}\sin x &= \sin \frac{{7\pi }}{6}\\x &= n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6} \qquad n \in Z\end{align}\]

Therefore, \(x = \left( {2n + 1} \right)\frac{\pi }{2}\) or \(n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6}\), where \(n \in Z\)

Chapter 3 Ex.3.4 Question 8

Find the principal and general solutions of the following equation:

\({\sec ^2}2x = 1 - \tan 2x\)

Solution

\[\begin{align}{\sec ^2}2x &= 1 - \tan 2x\\1 - {\tan ^2}2x &= 1 - \tan 2x\\{\tan ^2}2x - \tan 2x &= 0\\\tan 2x\left( {\tan 2x - 1} \right) &= 0\end{align}\]

Now,

\(\tan 2x = 0\)     or     \(\tan 2x - 1 = 0\)

Therefore,

\[\begin{align}\tan 2x &= 0\\2x &= n\pi \quad n \in Z\\x &= \frac{{n\pi }}{2}\end{align}\]

Or,

\[\begin{align}\tan 2x - 1 &= 0\\\tan 2x &= 1\\&= - \tan \frac{\pi }{4}\\&= \tan \left( {\pi - \frac{\pi }{4}} \right)\\&= \tan \frac{{3\pi }}{4}\\2x &= n\pi + \frac{{3\pi }}{4} \qquad n \in Z\\x &= \frac{{n\pi }}{2} + \frac{{3\pi }}{8}\end{align}\]

Therefore, \(x = \frac{{n\pi }}{2}\) or \(\left( \frac{n\pi }{2}+\frac{3\pi }{8} \right)\), where \(n\in Z\)

Chapter 3 Ex.3.4 Question 9

Find the principal and general solutions of the following equation:

\(\sin x + \sin 3x + \sin 5x = 0\)

Solution

\[\begin{align} & \sin x+\sin 3x+\sin 5x=0 \\ & 2\sin \left( \frac{5x+x}{2} \right)\cos \left( \frac{5x-x}{2} \right)+\sin 3x=0\\& \quad \qquad \left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ & 2\sin 3x\cos 2x+\sin 3x=0 \\ & \sin 3x\left( 2\cos 2x+1 \right)=0 \\ \end{align}\]

Now,

\(\sin 3x = 0\)     or     \(2\cos 2x + 1 = 0\)

Therefore,

\[\begin{align}\sin 3x &= 0\\3x &= n\pi \qquad n \in Z\\x &= \frac{{n\pi }}{3}\end{align}\]

Or,

\[\begin{align}2\cos 2x + 1 &= 0\\\cos 2x& = - \frac{1}{2}\\&= - \cos \frac{\pi }{3}\\&= \cos \left( {\pi - \frac{\pi }{3}} \right)\\&= \cos \frac{{2\pi }}{3}\\2x &= 2n\pi \pm \frac{{2\pi }}{3} \qquad n \in Z\\x &= n\pi \pm \frac{\pi }{3}\end{align}\]

Therefore, \(x = \frac{{n\pi }}{3}\) or \(\left( {n\pi \pm \frac{\pi }{3}} \right)\), where \(n \in Z\)

  
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