# 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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