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NCERT Solutions for Exercise 3.4 Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT Solutions for Exercise 3.4 Class 11 Maths Chapter 3 - Trigonometric Functions

Edited By Sumit Saini | Updated on Jul 29, 2022 03:20 PM IST

Mainly Principal and general solutions are discussed in the Exercise 3.4 Class 11 Maths trigonometric functions. NCERT Solutions for class 11 maths chapter 3 exercise 3.4 is the second last exercise in the sequence.most of the questions from this exercise can be seen in the board examination. Exercise 3.4 Class 11 Maths is a must to do exercise and it can help getting hold of some of the tricky concepts of this chapter. NCERT Solutions for class 11 maths chapter 3 exercise 3.4 is highly recommended to the students. Also for the other exercises of NCERT one can find the below.

This Story also Contains
  1. NCERT solutions for class 11 maths chapter 3 trigonometric function-Exercise: 3.4
  2. More About NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4

NCERT solutions for class 11 maths chapter 3 trigonometric function-Exercise: 3.4

Question:1 Find the principal and general solutions of the following equations: \tan x= \sqrt{3}

Answer:

It is given that given
\tan x= \sqrt{3}
Now, we know that \tan\frac{\pi}{3}= \sqrt3 and \tan\frac{4\pi}{3}= \tan \left ( \pi+\frac{\pi}{3} \right )=\sqrt3

Therefore,
the principal solutions of the equation are x = \frac{\pi}{3},\frac{4\pi}{3}
Now,
The general solution is \tan x =\tan \frac{\pi}{3}

x =n{\pi} + \frac{\pi}{3} where n \ \epsilon \ Z and Z denotes sets of integer

Therefore, the general solution of the equation is x =n{\pi} + \frac{\pi}{3} where n \ \epsilon \ Z and Z denotes sets of integer

Question:2 Find the principal and general solutions of the following equations: \small \sec x = 2

Answer:

We know that value of \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 principal solutions are x = \frac{\pi}{3} and \frac{5\pi}{3}
\sec x = \sec\frac{\pi}{3}
We know that value of sec x repeats after an interval of 2\pi
So, by this we can say that

the general solution is x = 2n\pi \pm \frac{\pi}{3} where n \epsilon Z

Question:3 Find the principal and general solutions of the following equations: \small \cot x = - \sqrt{3}

Answer:

we know that \ cot\frac{\pi}{6} = \sqrt{3} and we know that \ \cot\frac{5\pi}{6} = \cot\left ( \pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}

Similarly , the value for \ \cot\frac{11\pi}{6} = \cot\left ( 2\pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}
Therefore, principal solution is x = \frac{5\pi}{6} \ and \ \frac{11\pi}{6}

We also know that the value of cot x repeats after an interval of \pi
There the general solution is x = n\pi \pm \frac{5\pi}{6} \ where \ n \ \epsilon \ Z

Question:4 Find the principal and general solutions of the following equations: \small cosec x = -2

Answer:

We know that
cosec \frac{\pi}{6} = 2

cosec (\pi + \frac{\pi}{6}) = -cosec\frac{\pi}{6} = -2 and also cosec (2\pi - \frac{\pi}{6}) = cosec\frac{11\pi}{6} = -2
So,
cosec x= cosec\frac{7\pi}{6} and cosec x= cosec\frac{11\pi}{6}

So, the principal solutions are x = \frac{7\pi}{6} \ and \ \frac{11\pi}{6}

Now,
cosec x= cosec\frac{7\pi}{6}

\sin x = \sin\frac{7\pi}{6} \left ( \because \sin x = \frac{1}{cosec x} \right )

x = n\pi + (-1)^{n}\frac{7\pi}{6}
Therefore, the general solution is

x = n\pi + (-1)^{n}\frac{7\pi}{6}

where n \ \epsilon \ Z

Question:5 Find the general solution for each of the following equation \small \cos 4x = \cos 2x

Answer:

cos4x = cos2x
cos4x - cos2x = 0
We know that
\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
We use this identity
\therefore cos 4x - cos 2x = -2sin3xsinx
\Rightarrow -2sin3xsinx = 0 \Rightarrow sin3xsinx=0
So, by this we can that either
sin3x = 0 or sinx = 0
3x = n\pi x = n\pi
x = \frac{n\pi}{3} x = n\pi

Therefore, the general solution is

x=\frac{n\pi}{3}\ or\ n\pi \ where \ n\in Z

Question:6 Find the general solution of the following equation \small \cos 3x + \cos x -\cos 2x = 0

Answer:

We know that
\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\ and \\ \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
= cos2x(2cosx-1) = 0
So, either
cos2x = 0 or cosx=\frac{1}{2}
2x=(2n+1)\frac{\pi}{2} cosx =\cos\frac{\pi}{3}
x=(2n+1)\frac{\pi}{4} x =2n\pi \pm \frac{\pi}{3}

\therefore the general solution is

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

Question:7 Find the general solution of the following equation \small \sin 2x + \cos x = 0

Answer:

sin2x + cosx = 0
We know that
sin2x = 2sinxcosx
So,
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either

cosx = 0 or 2sinx + 1 = 0
x=(2n+1)\frac{\pi}{2} sinx =\sin\frac{7\pi}{6}
x=n\pi +(-1)^{n}\frac{7\pi}{6}
Therefore, the general solution is

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

Question:8 Find the general solution of the following equation \small \sec^{2}2x = 1 - \tan2x

Answer:

We know that
\sec^{2}x = 1 + \tan^{2}x
So,
1 + \tan^{2}2x = 1 -\tan2x
\tan^{2}2x + \tan2x = 0\\ \\ \tan2x(\tan2x+1) = 0
either
tan2x = 0 or tan2x = -1 ( \tan x = \tan \left ( \pi - \frac{\pi}{4} \right ) = \tan\frac{3\pi}{4} )
2x = n\pi 2x=n\pi + \frac{3\pi}{4}
x=\frac{n\pi}{2} x=\frac{n\pi}{2} + \frac{3\pi}{8}
Where n \epsilon Z

Question:9 Find the general solution of the following equation \small \sin x + \sin 3x + \sin 5x = 0

Answer:

We know that
\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
We use this identity in our problem
\sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} =2\sin3x\cos2x
Now our problem simplifeis to
2\sin3x\cos2x+ \sin3x = 0
take sin3x common
\sin3x(2\cos2x+ 1) = 0
So, either
sin3x = 0 or \cos2x = -\frac{1}{2} \left ( \cos2x = -\cos\frac{\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\frac{2\pi}{3} \right )
3x = n\pi 2x = 2n\pi \pm \frac{2\pi}{3}
x = \frac{n\pi}{3} x = n\pi \pm \frac{\pi}{3}
Where n \ \epsilon \ Z

More About NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4

The NCERT class 11 maths chapter Trigonometric functions deals with the basic trigonometric functions. Exercise 3.4 Class 11 Maths is a good source for practice and scoring good marks. As far as final exam for class 11 is concerned the chance of getting a question from the Class 11 maths chapter 3 exercise 3.4 is high. Identities related to sum and difference of two angles and its problems are discussed in the exercise 3.4 class 11 maths

Benefits of NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4

  • The Class 11th maths chapter 3 exercise is also helpful in physics also.

  • Exercise 3.4 Class 11 Maths needs only practice for one time. If done properly, a good score can be achieved.

  • Class 11 maths chapter 3 exercise 3.4 solutions are provided here for the students which make them easier to understand in comprehensive manner.

Also see-

NCERT solutions of class 11 subject wise

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Subject wise NCERT Exampler solutions

Frequently Asked Question (FAQs)

1. What is the difficulty level of this exercise?

Some of the questions are difficult but it is manageable after attempting previous exercises

2. What are the general methods used to solve these questions ?

Students need to use only methods which are there in the previous exercises. For more information they can refer to the solutions provided above.

3. How scoring is this chapter in the CBSE exam ?

Approx 2-3 questions are asked from this chapter.

4. How much time will it take to cover exercise 3.4 maths class 11 ?

4-5 hours dedicatedly is sufficient.

5. How many questions are there in exercise 3.4 class 11 maths ?

There are 9 questions in total

6. Which concepts are there mostly in exercise 3.4 maths CBSE class 11 ?

Concepts related to finding principal and general solutions are discussed mainly.

7. Is a 100% score achievable in this exercise in Board exam ?

Through pracise full score can be easily achieved.

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