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Exercise 14 (Solutions) Textbook of lgebra and Trigonometry for Class XI

Available online @ http://www.mathcity.org , Version: 1.1.4

Question # 1

(i ) Since3

sin2

x = - 1 3sin2

x - æ ö

Þ = -ç ÷

è ø

5 4,

3 3 x

p p Þ = where [ ]0, 2 x p Î

(ii) Since cosec 2q = 1

2sin q

Þ = 1sin2

q Þ =

1 1sin2

q - æ öÞ = ç ÷è ø

5

,6 6

p p = where [ ]0, 2q p Î

(iii) Do yourself(iv) Since

1cot

3q =

1 1tan 3q

Þ = tan 3q Þ =

( )1tan 3q -Þ = 4,3 3p p

q Þ = where [ ]0, 2q p Î

Question # 2 (i)

Since 2 1tan3

q = 1tan3

q Þ = ±

1tan

3q Þ = or 1tan

3q = -

1 1tan3


or 11

tan3

q - æ ö= -ç ÷è ø

6

p q Þ = or 5

6

p q =

Since period of tan q is p Therefore general value of q =

6n

p p + , 5

6n

p p +

So Solution Set5

6 6n n

p p p p

ì ü ì ü= + +í ý í ýî þ î þ

U where n Î ¢

Question # 2 (ii )

Since 24

cosec3

q = 2cosec3

q Þ = ±

2cosec3

q Þ = or 2cosec3

q = -

3sin

2q Þ = or 3sin

2q = -

1 3sin2

q - æ ö

Þ = ç ÷è ø

or 13

sin2

q - æ ö

= -ç ÷è ø

2,

3 3

p p q Þ = or 4 5,

3 3

p p q =

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Sc I / Ex. 14 - 2

Since period of cosec q is 2p

Therefore general value of q =2 4 5

2 , 2 , 2 , 23 3 3 3

n n n np p p p

p p p p + + + +

Solution set =2 4 5

2 2 2 23 3 3 3

n n n np p p p

p p p p ì ü ì ü ì ü ì ü+ + + +í ý í ý í ý í ýî þ î þ î þ î þ

U U U where n Î ¢ .

Question # 2 (ii i)Since 2

4sec

3q = 2sec

3q Þ = ±

2sec

3q Þ = or 2sec

3q = -

3cos

2q Þ = or 3cos

2q = -

1 3cos2

q - æ ö

Þ = ç ÷è ø

or 13

cos2

q - æ ö

= -ç ÷è ø

11,

6 6

p p q Þ = or 5 7,

6 6

p p q =

Q period of sec q is 2p

\ general values of 5 7 112 , 2 , 2 , 26 6 6 6

n n n np p p p

q p p p p = + + + +

S.Set5 7 11

2 2 2 26 6 6 6

n n n np p p p

p p p p ì ü ì ü ì ü ì ü= + + + +í ý í ý í ý í ýî þ î þ î þ î þ


Question # 2 (iv) Do yourself

Question # 323tan 2 3 tan 1 0q q + + =

( ) ( )( ) ( )2 2

3 tan 2 3 tan 1 1 0q q Þ + + =

( )2

3 tan 1 0q Þ + = ( )3 tan 1 0q Þ + = 3 tan 1q Þ = -

1tan3

q Þ = - 1 1tan 3q - æ öÞ = -ç ÷è ø 56

p q Þ =

Q period of tan q is p

\ general value of 56

np

q p = + , n Î ¢

Question # 42tan sec 1 0q q - - =

( )2sec 1 sec 1 0q q Þ - - - = 2sec 1 sec 1 0q q Þ - - - = 2sec sec 2 0q q Þ - - = 2sec 2sec sec 2 0q q q Þ - + - = ( ) ( )sec sec 2 1 sec 2 0q q q Þ - + - =

( )( )sec 1 sec 2 0q q Þ + - = ( )sec 1 0q Þ + = or ( )sec 2 0q - = sec 1q Þ = - or sec 2q = +

cos 1q Þ = - or 1cos2

q =

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Sc I / Ex. 14 - 3

( )1cos 1q -Þ = - or 1 1cos2

q - æ ö= ç ÷è ø

32

p q Þ = or 5,

6 6

p p q =

Q period of cos q is 2p

\ general value of q = 3 52 , 2 , 22 6 6

n n np p p

p p p + + + where n Î ¢

Question # 522sin cos 1 0q q + - =

22sin 1 sin 1 0q q Þ + - - = 2sin 2sin 0q q Þ - + = ( )sin sin 2 0q q Þ - - =

sin 0q Þ - = or sin 2 0q - = sin 0q Þ = or sin 2q =

( )1

sin 0q -Þ =

Which does not hold as sin [ 1,1]q Î -

0 ,q p Þ = Q period of sin q is 2p \ general value of q 0 2 , 2n np p p = + +

= 2 , 2n np p p + where n Î ¢

Question # 62 23cos 2 3sin cos 3sin 0q q q q - - =

Dividing throughout by 2cos q 2 2

2 2 23cos 2 3sin cos 3sin 0cos cos cos

q q q q

q q q - - =

23 2 3 tan 3tan 0q q Þ - - = 23tan 2 3 tan 3 0q q Þ - - + =

23tan 2 3 tan 3 0q q Þ + - = ×ing by -1

( ) ( )( )( )

2

2 3 2 3 4 3 3tan

2 3q

- ± - -Þ =

2 3 12 36tan

6q

- ± +Þ = 2 3 486

- ±=

2 3 16 36

- ± ´= 2 3 4 36

- ±=

2 3 4 3tan

6q

- +Þ = 2 36

= or 2 3 4 3tan6

q - -= 6 3

6= -

3tan

3

q Þ =

( )2

3

3

= 1

3

= or tan 3q Þ = -

1 1tan3


or ( )1tan 3q -= -

6

p q Þ = or 11

6

p q =

Q period of tan q is p

\ general value of q 6

np

p = + , 116

np

p + where n Î ¢ .

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Sc I / Ex. 14 - 4

Question # 824sin 8cos 1 0q q - + =

( )24 1 cos 8cos 1 0q q Þ - - + = 24 4cos 8cos 1 0q q Þ - - + = 24cos 8cos 5 0q q Þ - - + = 24cos 8cos 5 0q q Þ + - =

24cos 10cos 2cos 5 0q q q Þ + - - =

( ) ( )2cos 2cos 5 1 2cos 5 0q q q Þ + - + =

( )( )2cos 5 2cos 1 0q q Þ + - =

2cos 5 0q Þ + = or 2cos 1 0q - = 2cos 5q Þ = - or 2cos 1q =

5cos

2q

-Þ = or 1cos2

q =

1 5cos2

q - -æ öÞ = ç ÷è ø

or 11

cos2

q - æ ö= ç ÷è ø

Which is not possible as cos [ 1,1]q Î - or 5,

3 3

p p q =


\ general value of 52 , 23 3

n np p

q p p = + + where n Î ¢ .

Question # 93 tan sec 1 0 x x- - = ………… ( )i

sin 13 1 0

cos cos x x x

Þ - - =

3 sin 1 cos 0 x xÞ - - = ´ ing by cos q .3sin 1 cos x xÞ - =

On squaring both sides.

( ) ( )2 2

3 sin 1 cos x x- = 2 23sin 2 3 sin 1 cos x x xÞ - + = 2 23sin 2 3 sin 1 1 sin x x xÞ - + = - 2 23sin 2 3sin 1 1 sin 0 x x xÞ - + - + = 24sin 2 3 sin 0 x xÞ - =

( )2sin 2sin 3 0 x xÞ - = 2sin 0 xÞ = or 2sin 3 x =

sin 0Þ = or 3sin2

x =

( )1sin 0 x -Þ = or 1 3sin2

x - æ ö

= ç ÷è ø

0 xÞ = , p or3

x p = , 2

3

p

Now to check extraneous roots put 0 x = in ( )i L.H.S = 3 tan(0) sec(0) 1- - 0 1 1= - - 2= - 0¹ = R.H.S

Implies that 0 x = is an extraneous root of given equation. Now put x p = in ( i)

L.H.S = 3 tan( ) sec( ) 1p p - - 0 ( 1) 1= - - - 0= = R.H.SImplies that x p = is a root of the equation.

Now put3

x p = in ( )i

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Sc I / Ex. 14 - 5

L.H.S = 3 tan sec 13 3

p p æ ö æ ö- -ç ÷ ç ÷è ø è ø ( )3 3 2 1= - - 1 2 1= - - 0= = R.H.S

Implies that3

x p = is a root of given equation.. Since period of tan is p .

Now put23

x p = in ( )i

L.H.S = 2 23 tan sec 13 3p p æ ö æ ö- -ç ÷ ç ÷è ø è ø

( ) ( )3 3 2 1= - - - - 3 2 1= - + - 2= - = R.H.SImplies

23

x p = is an extraneous root of given equation.

Q period of sin is 2 p

\ general values of 2 x np p = + , 23

np

p +

Solution Set = { }2 23

n np

p p p ì ü+ +í ýî þ

U where n Î ¢ .

Question # 10cos2 sin 3 x x=

2 2 3cos sin 3sin 4sin x x x xÞ - = - 2 2 3cos sin 3sin 4sin 0 x x x xÞ - - + =

( )2 2 31 sin sin 3sin 4sin 0 x x x xÞ - - - + = 2 31 2sin 3sin 4sin 0 x x xÞ - - + =

3 24sin 2sin 3sin 1 0 x x xÞ - - + = Take sin 1 x = as a root then by synthetic division

( )( )2sin 1 4sin 2sin 1 0 x x xÞ - + - = sin 1 0 xÞ - = or 24sin 2sin 1 0 x x+ - =

sin 1 xÞ = or( ) ( )( )

( )

22 2 4 4 1

sin2 4

x- ± - -

=

1sin (1) x -Þ = or 2 4 16sin8

x - ± += 2 20

8

- ±=

2 x

p Þ = or 2 20sin8

x - += or 2 20sin

8 x

- -=

sin 0.309 x = or sin 0.809= - ( )1sin 0.309 x -Þ = or ( )1sin 0.809 x -= -

18,162» or 234 , 306»

18 , 162180 180

x p p Þ = × × or 234 , 306

180 180 x

p p = × ×

9,

10 10

p p = or 13 17,10 10

p p =

1 4 -2 -3 1¯ 4 2 -14 2 -1 0

Q 2 2cos2 cos sin x x x= - 3sin3 3sin 4sin x x x= -

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Sc I / Ex. 14 - 6

Q period of sin is 2 p

\ general value of 9 13 172 , 2 , 2 , 2 , 210 10 10 10 2

n n n n np p p p p

p p p p p = + + + + +

S.Set9 13 17

2 2 2 2 210 10 10 10 2

n n n n np p p p p

p p p p p ì ü ì ü ì ü ì ü ì ü= + + + + +í ý í ý í ý í ý í ýî þ î þ î þ î þ î þ

U U U U

Question No. 11 sec3 secq q = 1 1

cos3 cosq q Þ =

cos3 cosq q Þ = 34cos 3cos cosq q q Þ - = 3cos3 4cos 3cosq q q = -Q 34cos 3cos cos 0q q q Þ - - = 34cos 4cos 0q q Þ - =

( )2

4cos cos 1 0q q Þ - = 4cos 0q Þ = or 2cos 1 0q - = cos 0q Þ = or 2cos 1q =

1cos (0)q -Þ = or cos 1q = ± ( ) ( )1 1cos 1 , cos 1q q - -Þ = = - 3

,2 2

p p q Þ = or 0 ,q p =


\ general values of 32 , 2 , 0 2 , 22 2

n n n np p

q p p p p p = + + + +

32 , 2 ,

2 2n n n

p p p p p = + +

S. Set { }32 22 2

n n np p

p p p ì ü ì ü= + +í ý í ýî þ î þ

U U where n Î ¢ .

Question # 12tan 2 cot 0q q + = tan 2 cotq q Þ = -

sin 2 coscos2 sin

q q q q

Þ = - 2 22sin cos coscos sin sinq q q

q q q Þ = -

-

( )( ) ( )( )2 22sin cos sin cos cos sinq q q q q Þ = - - 2 3 22sin cos cos sin cosq q q q q Þ = - + 2 3 22sin cos cos sin cos 0q q q q q Þ + - =

2 3sin cos cos 0q q q Þ + =

( )2 2cos sin cos 0q q q Þ + = ( )cos 1 0q Þ = 1

cos (0)q -

Þ = 3

,2 2

p p =


\ general values of 32 , 22 2

n np p

q p p = + +

S. Set =3

2 22 2

n np p

p p ì ü ì ü+ +í ý í ýî þ î þ

U where n Î ¢ .

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Sc I / Ex. 14 - 7

Question # 13sin 2 sin 0 x x+ =

2sin cos sin 0 x xÞ + = sin 2 2sin cosq q q =Q ( )sin 2cos 1 0 x xÞ + =

sin 0 xÞ = or 2cos 1 0 x + = 1

sin (0) x -Þ =

or1

cos 2 x = -

1 1

cos 2 x - æ öÞ = -

ç ÷è ø

0 , x p Þ = or 2 4,3 3

x p p =

Q period of sin x and cos is 2 p

\ general values of 2 40 2 , 2 , 2 , 23 3

x n n n np p

p p p p p = + + + +

2 4, 2 , 2

3 3n n n

p p p p p = + +

So solution set { } 2 42 23 3

n n np p p p p ì ü ì ü= + +í ý í ýî þ î þ

U U where n Î ¢ .

Question # 14sin 4 sin 2 cos3 x x- =

4 2 4 22cos sin cos3

2 2 x x x x

x+ -æ ö æ öÞ =ç ÷ ç ÷è ø è ø

2cos3 sin cos3 0 x xÞ - = ( )cos3 2sin 1 0 x xÞ - =

cos3 0 xÞ = or 2sin 1 0 x - = 13 cos (0) x -Þ = , 1sin

2 x =

33 ,

2 2 x

p p Þ = , 1 1sin2

x - æ ö= ç ÷è ø

,6 2

x p p Þ = , 5,

6 6 x

p p =

Since period of cos3 x is

2

3

p

and period of sin is 2p

\ general values of 2 2 5, , 2 , 26 3 2 3 6 6

n n x n n

p p p p p p p p = + + + +

So solution set2 2 5

2 26 3 2 3 6 6

n nn n

p p p p p p p p

ì ü ì ü ì ü ì ü= + + + +í ý í ý í ý í ýî þ î þ î þ î þ


Question # 15sin cos3 cos5 x x x+ = sin cos5 cos3 x x xÞ = -

5 3 5 3sin 2sin sin

2 2 x x x x

x + -æ ö æ öÞ = - ç ÷ ç ÷è ø è ø

sin 2sin 4 sin x x xÞ = - sin 2sin 4 sin 0 x x xÞ + =

( )sin 1 2sin 4 0 x xÞ + = sin 0Þ = or 1 2sin 4 0 x+ =

1sin (0) x -Þ = or 1sin42

x = - 1 14 sin2

x - æ öÞ = -ç ÷è ø

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Sc I / Ex. 14 - 8

0 , x p Þ = or 7 114 ,6 6

x p p = 7 11,

24 24 x

p p Þ =

Since period of sin is 2 p and period of sin 4 x is24 2

p p =

\ general values of 7 110 2 , 2 , ,24 2 24 2

n n x n n

p p p p p p p = + + + +

So solution set { } { } 7 112 224 2 24 2

n nn n p p p p p p p ì ü ì ü= + + +í ý í ýî þ î þ


Question # 16sin3 sin 2 sin 0 x x+ + =

( )sin3 sin sin 2 0 x x xÞ + + = 3 3

2sin cos sin 2 02 2

x x x x x

+ -æ ö æ öÞ + =ç ÷ ç ÷è ø è ø 2sin 2 cos sin 2 0 x x xÞ + =

( )sin 2 2cos 1 0 x xÞ + =

sin 2 0 xÞ = or 2cos 1 0 x + = 12 sin (0) x -Þ = or 1cos

2 x = -

2 0 , x p Þ = or 1 1cos2

x - æ ö= -ç ÷è ø

0 ,2

x p Þ = or 2 4,

3 3 x

p p =

Since period of sin 2 x is22

p

p = and period of cos x is 2p

\ general values of 2 40 , , 2 , 22 3 3

x n n n np p p

p p p p = + + + +

S. Set { } 2 42 22 3 3

n n n np p p

p p p p ì ü ì ü ì ü= + + +í ý í ý í ýî þ î þ î þ


Question # 17sin 7 sin sin3 x x x- =

7 72cos sin sin 32 2

x x x x x

+ -æ ö æ öÞ =ç ÷ ç ÷è ø è ø

2cos4 sin 3 sin 3 0 x x xÞ - = ( )sin 3 2cos4 1 0 x xÞ - =

sin 3 0 xÞ = or 2cos4 1 0 x - =

( )13 sin 0 x -Þ = or 1cos42

x =

3 0 , p Þ = or 1 1 54 cos ,2 3 3

x p p - æ ö= =ç ÷è ø

0 ,3

x p Þ = or 5,

12 12 x

p p =

Since period of sin3 is23

p and period of cos4 x is

24 2

p p =

\ general values of 2 2 50 , , ,3 3 3 12 2 12 2n n n n

x p p p p p p p = + + + +

So S. set2 2 5

3 3 3 12 2 12 2n n n np p p p p p p ì ü ì ü ì ü ì ü= + + +í ý í ý í ý í ý

î þ î þ î þ î þU U U where n Î ¢ .

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Sc I / Ex. 14 - 9

Question # 18sin sin3 sin5 0 x x+ + = ( )sin5 sin sin3 0 x x xÞ + + =

5 52sin cos sin3 0

2 2 x x x x

x+ -æ ö æ öÞ × + =ç ÷ ç ÷è ø è ø

2sin 3 cos2 sin3 0 x x xÞ + = ( )sin 3 2cos 2 1 0 x xÞ + =

sin 3 0 xÞ = or 2cos2 1 0 x + =

( )13 sin 0 x -Þ = or 2cos2 1 x = - 1 12 cos2

x - æ öÞ = -ç ÷è ø

3 0 , p Þ = or 2 42 ,3 3

x p p =

0 ,3

x p Þ = or 2,

3 3 x

p p =

Since period of sin3 x is23

p and period of cos2 x is

22

p p =

\ general values of 2 2 20 , , ,3 3 3 3 3n n

n np p p p p

p p = + + + +

S.Set =2 2 2

3 3 3 3 3n n

n np p p p p

p p ì ü ì ü ì ü ì ü+ + +í ý í ý í ý í ýî þ î þ î þ î þ

U U U where n Î Z

Question # 19

sin sin3 sin5 sin 7 0q q q q

+ + + = ( ) ( )sin 7 sin sin 5 sin3 0q q q q Þ + + + = 7 7 5 3 5 3

2sin cos 2sin cos 02 2 2 2

q q q q q q q q + - + -æ ö æ ö æ ö æ öÞ + =ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø

2sin 4 cos3 2sin 4 cos 0q q q q Þ + = ( )2sin 4 cos3 cos 0q q q Þ + =

3 32sin 4 2cos cos 0

2 2

q q q q q

æ + - öæ ö æ öÞ =ç ÷ ç ÷ç ÷è ø è øè ø

4sin 4 cos 2 cos 0q q q Þ = sin 4 0q Þ = or cos2 0q = or cos 0q =

( )14 sin 0q -Þ = , ( )12 cos 0q -= , ( )1cos 0q -=

4 0 ,q p Þ = or 32 ,2 2

p p q = or 3,

2 2

p p q =

0 ,4

p q Þ = or 3,

4 4

p p q =

Since period of sin 4 q is2

4 2

p p = , cos2 q is 2

2

p p = and cos q is 2p

\ general values of 3 30 , , , , 2 , 22 4 2 4 4 2 2

n nn n n n

p p p p p p p q p p p p = + + + + + +

S. Set3 3

2 22 4 2 4 4 2 2

n nn n n n

p p p p p p p p p p p

ì ü ì ü ì ü ì ü ì ü ì ü= + + + + +í ý í ý í ý í ý í ý í ýî þ î þ î þ î þ î þ î þ

U U U U U .

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Sc I / Ex. 14 - 10

Question # 20cos cos3 cos5 cos7 0q q q q + + + = ( ) ( )cos7 cos cos5 cos3 0q q q q Þ + + + =

7 7 5 3 5 32cos cos 2cos cos 0

2 2 2 2

q q q q q q q q + - + -æ ö æ ö æ ö æ öÞ + =ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø

2cos4 cos3 2cos4 cos 0q q q q Þ + =

( )2cos4 cos3 cos 0q q q Þ + = 3 3

2cos4 2cos cos 02 2

q q q q q

æ + - öæ ö æ öÞ =ç ÷ ç ÷ç ÷è ø è øè ø

4cos4 cos2 cos 0q q q Þ = Now do yourself as above question.

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