Class 11 Maths Ncert Solutions Chapter 3 Ex 3.4 Trigonometric Functions PDF

Class 11 Maths Ncert Solutions Ex 3.4

Class 11 Maths Ncert Solutions Chapter 3 Ex 3.4

 Q.1: Find general solutions and the principle solutions of the given equation: tan x = 3

 

Sol:

tan x = 3    [Given]

As we know that, tanπ3=3

And, tan4π3 = tan(π+π3) = tan(π3) = 3

Therefore, the principle solutions are x=π3 and 4π3

Now, tanx=tanπ3

x=nπ+π3, where n Z

Therefore, the general solution is x=nπ+π3, where n Z.

 

 

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

 

Sol:

sec x = 2   [Given]

As we know that, secπ3=2

And, sec5π3 = sec(2ππ3) = secπ3 = 2

Therefore, the principle solutions are x=π3 and 5π3.

Now, secx=secπ3

And, cosx=cosπ3[secx=1cosx]

x=2nπ±π3, where n Z

Therefore, the general solution is x=2nπ±π3, where n Z.

 

 

Q.3: Find general solutions and the principle solutions of the given equation:

cot = 3

 

Sol:

cot = 3          [Given]

As we know that, cotπ6=3cot(ππ6)=cotπ6=3

And, cot(2ππ6) = cotπ6 = 3

That is cot5π6=3 and cot11π6=3

Therefore, the principle solutions are x=5π6 and 11π6.

Now,cotx=cot5π6

And, tanx=tan5π6[cotx=1tanx]

x=nπ+5π6, where n Z

Therefore, the general solution is x=nπ+5π6, where n Z.

 

 

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

 

Sol:

cosec x = -2     [Given]

As we know that, cosecπ6=2

Hence, cosec(π+π6) = cosecπ6 = -2

And, cosec(2ππ6) = cosecπ6 = -2

That is cosec7π6=2 and cosec11π6=2.

Therefore, the principle solutions are x=7π6 and 11π6.

Now,cosecx=cosec7π6

And, sinx=sin7π6[cosecx=1sinx]

x=nπ+(1)n7π6, where n Z

Therefore, the general solution is x=nπ+(1)n7π6, where n Z.

 

 

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Sol:

cos 4x = cos 2x           [Given]

i.e.  cos 4x – cos 2x = 0

2sin(4x+2x2)sin(4x2x2)=0

Since, cos A – cos B = 2sin(A+B2)sin(AB2)

(sin 3x)  (sin x) = 0

sin 3x = 0  or  sin x = 0

sin 3x = 0

3x=nπ

x=nπ3, where n Z

sin x = 0

  x=nπ, where n Z

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Sol:

cos 3x + cos x – cos 2x = 0        [Given]

2cos(3x+x2)cos(3xx2)cos2x=0

Since, cos A + cos B = 2cos(A+B2)cos(AB2)

2cos 2x cos x – cos 2x = 0

cos 2x (2cos x – 1) = 0

cos 2x = 0  or  2cos x -1 = 0

cos 2x = 0

2x=(2n+1)π2, where n Z

x=(2n+1)π4, where n Z

2cos x -1 = 0

cosx=12

cosx=cosπ3, where n Z

x=2nπ±π3, where n Z

 

 

Q.7: Find the general solution of the given equation:  sin 2x + cos x = 0

 

Sol:

sin 2x + cos x = 0          [Given]

2sin x cos x + cos x = 0

cos x (2sin x + 1) = 0

cos x = 0   or   2sin x + 1 = 0

cos x = 0

cosx=(2n+1)π2, where n Z

2sin x + 1 = 0

= sinπ6 = sin(π+π6) = sin7π6

  x=nπ+(1)n7π6, where n Z

Therefore, the general solution is (2n+1)π2 or nπ+(1)n7π6, where n Z.

 

 

Q.8: Find the general solution of the given equation:

sec22x=1tan2x

 

Sol:

sec22x=1tan2x           [Given]

1+tan22x=1tan2x tan22x+tan2x=0

tan 2x ( tan 2x + 1) = 0

tan 2x = 0  or  tan 2x + 1 = 0

tan 2x = 0

tan 2x = tan 0

2x = n∏ + 0, where n Z

x=nπ2, where n Z

tan 2x + 1 = 0

tan 2x = -1= tanπ4= tan(ππ4)= tan3π4

  2x=nπ+3π4, where n Z

  x=nπ2+3π8, where n Z

Therefore, the general solution is nπ2 or nπ2+3π8, where n Z.

 

 

Q.9: Find the general solution of the given equation:  sin x + sin 3x + sin 5x = 0

 

Sol:

sin x + sin 3x + sin 5x = 0              [Given]

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

[2sin(x+5x2)cos(x5x2)]+sin3x=0

Since, sin A + sin B = 2sin(A+B2)cos(AB2)

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

2sin 3x cos 2x + sin 3x = 0

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

sin 3x = 0 or  2cos 2x + 1 = 0

Now,

sin 3x = 0

3x=nπ, where n Z

x=nπ3, where n Z

2cos 2x + 1 = 0

cos2x=12 = cosπ3 = cos(ππ3) = cos2π3

2x=2nπ±2π3, where n Z

x=nπ±π3, where n Z

Therefore, the general solution is nπ3 or nπ±π3, where n Z.

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