# RD Sharma Solutions for Class 11 Maths Chapter 11 Trigonometric Equations

## RD Sharma Solutions Class 11 Maths Chapter 11 â€“ Get Free PDF (for 2021-22)

RD Sharma Solutions for Class 11 Maths Chapter 11 – Trigonometric EquationsÂ is given here for students to score good marks in the board exams. This chapter gives an overview of Trigonometric Equations (the equations containing trigonometric functions of unknown angles are known as trigonometric equations). For students who face difficulty in understanding the concepts during class hours, the experts at BYJUâ€™S have developed the solutions based on the studentsâ€™ grasping abilities.Â RD Sharma Solutions strictly follows the latest CBSE syllabus directed by the CBSE board.Â

Students can solve both chapter-wise and exercise-wise problems to increase their confidence level before appearing for the board exam. To boost interest among students in this chapter, RD Sharma Solutions for Class 11 Chapter 11 Trigonometric Equations pdf links are given here for easy access.

Chapter 11 – Trigonometric Equations contain one exercise and the RD Sharma Solutions present in this page provide solutions to the questions present in this exercise. Now, let us have a look at the concepts discussed in this chapter.

• Some definitions.
• General solutions of trigonometric equations.
• General solutions of trigonometric equations in specific forms.

## Download the Pdf of RD Sharma Solutions for Class 11 Maths Chapter 11 – Trigonometric Equations

### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 11 – Trigonometric Equations

1. Find the general solutions of the following equations:

(i) sin x = 1/2

(ii) cos x = – âˆš3/2

(iii) cosec x = – âˆš2

(iv) sec x = âˆš2

(v) tan x = -1/âˆš3

(vi) âˆš3 sec x = 2

Solution:

The general solution of any trigonometric equation is given as:

sin x = sin y, implies x = nÏ€ + (â€“ 1)n y, where nÂ âˆˆÂ Z.

cos x = cos y, implies x = 2nÏ€Â Â±Â y, where nÂ âˆˆÂ Z.

tan x = tan y, implies x = nÏ€Â + y, where nÂ âˆˆÂ Z.

(i) sin x = 1/2

We know sin 30o = sin Ï€/6 = Â½

So,

Sin x = sin Ï€/6

âˆ´ the general solution is

x = nÏ€ + (â€“ 1) n Ï€/6, where nÂ âˆˆÂ Z. [since, sin x = sin A => x = nÏ€ + (â€“ 1) n A]

(ii) cos x = – âˆš3/2

We know, cos 150o = (- âˆš3/2) = cos 5Ï€/6

So,

Cos x = cos 5Ï€/6

âˆ´ the general solution is

x = 2nÏ€Â Â± 5Ï€/6, where nÂ ÏµÂ Z.

(iii) cosec x = – âˆš2

Let us simplify,

1/sin x = – âˆš2 [since, cosec x = 1/sin x]

Sin x = -1/âˆš2

= sin [Ï€ + Ï€/4]

= sin 5Ï€/4 or sin (-Ï€/4)

âˆ´ the general solution is

x = nÏ€ + (-1)n+1 Ï€/4, where nÂ ÏµÂ Z.

(iv) sec x = âˆš2

Let us simplify,

1/cos x = âˆš2 [since, sec x = 1/cos x]

Cos x = 1/âˆš2

= cos Ï€/4

âˆ´ the general solution is

x = 2nÏ€Â Â± Ï€/4, where nÂ ÏµÂ Z.

(v) tan x = -1/âˆš3

Let us simplify,

tan x = -1/âˆš3

tan x = tan (Ï€/6)

= tan (-Ï€/6) [since, tan (-x) = -tan x]

âˆ´ the general solution is

x = nÏ€ + (-Ï€/6), where nÂ ÏµÂ Z.

or x = nÏ€ – Ï€/6, where nÂ ÏµÂ Z.

(vi) âˆš3 sec x = 2

Let us simplify,

sec x = 2/âˆš3

1/cos x = 2/âˆš3

Cos x = âˆš3/2

= cos (Ï€/6)

âˆ´ the general solution is

x = 2nÏ€Â Â± Ï€/6, where nÂ ÏµÂ Z.

2. Find the general solutions of the following equations:

(i) sin 2x = âˆš3/2

(ii) cos 3x = 1/2

(iii) sin 9x = sin x

(iv) sin 2x = cos 3x

(v) tan x + cot 2x = 0

(vi) tan 3x = cot x

(vii) tan 2x tan x = 1

(viii) tan mx + cot nx = 0

(ix) tan px = cot qx

(x) sin 2x + cos x = 0

(xi) sin x = tan x

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

Solution:

The general solution of any trigonometric equation is given as:

sin x = sin y, implies x = nÏ€ + (â€“ 1)n y, where nÂ âˆˆÂ Z.

cos x = cos y, implies x = 2nÏ€Â Â±Â y, where nÂ âˆˆÂ Z.

tan x = tan y, implies x = nÏ€Â + y, where nÂ âˆˆÂ Z.

(i) sin 2x = âˆš3/2

Let us simplify,

sin 2x = âˆš3/2

= sin (Ï€/3)

âˆ´ the general solution is

2x = nÏ€ + (-1)n Ï€/3, where nÂ ÏµÂ Z.

x = nÏ€/2 + (-1)n Ï€/6, where nÂ ÏµÂ Z.

(ii) cos 3x = 1/2

Let us simplify,

cos 3x = 1/2

= cos (Ï€/3)

âˆ´ the general solution is

3x = 2nÏ€ Â± Ï€/3, where nÂ ÏµÂ Z.

x = 2nÏ€/3 Â± Ï€/9, where nÂ ÏµÂ Z.

(iii) sin 9x = sin x

Let us simplify,

Sin 9x â€“ sin x = 0

Using transformation formula,

Sin A â€“ sin B = 2 cos (A+B)/2 sin (A-B)/2

So,

= 2 cos (9x+x)/2 sin (9x-x)/2

=> cos 5x sin 4x = 0

Cos 5x = 0 or sin 4x = 0

Let us verify both the expressions,

Cos 5x = 0

Cos 5x = cos Ï€/2

5x = (2n + 1)Ï€/2

x = (2n + 1)Ï€/10, where nÂ ÏµÂ Z.

sin 4x = 0

sin 4x = sin 0

4x = nÏ€

x = nÏ€/4, where nÂ ÏµÂ Z.

âˆ´ the general solution is

x = (2n + 1)Ï€/10 or nÏ€/4, where nÂ ÏµÂ Z.

(iv) sin 2x = cos 3x

Let us simplify,

sin 2x = cos 3x

cos (Ï€/2 â€“ 2x) = cos 3x [since, sin A = cos (Ï€/2 â€“ A)]

Ï€/2 â€“ 2x = 2nÏ€ Â± 3x

Ï€/2 â€“ 2x = 2nÏ€ + 3x [or] Ï€/2 â€“ 2x = 2nÏ€ – 3x

5x = Ï€/2 + 2nÏ€ [or] x = 2nÏ€ – Ï€/2

5x = Ï€/2 (1 + 4n) [or] x = Ï€/2 (4n – 1)

x = Ï€/10 (1 + 4n) [or] x = Ï€/2 (4n – 1)

âˆ´ the general solution is

x = Ï€/10 (4n + 1) [or] x = Ï€/2 (4n – 1), where nÂ ÏµÂ Z.

(v) tan x + cot 2x = 0

Let us simplify,

tan x = – cot 2x

tan x = – tan (Ï€/2 â€“ 2x) [since, cot A = tan (Ï€/2 – A)]

tan x = tan (2x – Ï€/2) [since, – tan A = tan -A]

x = nÏ€ + 2x – Ï€/2

x = nÏ€ – Ï€/2

âˆ´ the general solution is

x = nÏ€ – Ï€/2, where nÂ ÏµÂ Z.

(vi) tan 3x = cot x

Let us simplify,

tan 3x = cot x

tan 3x = tan (Ï€/2 – x) [since, cot A = tan (Ï€/2 – A)]

3x = nÏ€ + Ï€/2 â€“ x

4x = nÏ€ + Ï€/2

x = nÏ€/4 + Ï€/8

âˆ´ the general solution is

x = nÏ€/4 + Ï€/8, where nÂ ÏµÂ Z.

(vii) tan 2x tan x = 1

Let us simplify,

tan 2x tan x = 1

tan 2x = 1/tan x

= cot x

tan 2x = tan (Ï€/2 – x) [since, cot A = tan (Ï€/2 – A)]

2x = nÏ€ + Ï€/2 â€“ x

3x = nÏ€ + Ï€/2

x = nÏ€/3 + Ï€/6

âˆ´ the general solution is

x = nÏ€/3 + Ï€/6, where nÂ ÏµÂ Z.

(viii) tan mx + cot nx = 0

Let us simplify,

tan mx + cot nx = 0

tan mx = – cot nx

= – tan (Ï€/2 – nx) [since, cot A = tan (Ï€/2 – A)]

tan mx = tan (nx + Ï€/2) [since, – tan A = tan -A]

mx = kÏ€ + nx + Ï€/2

(m – n) x = kÏ€ + Ï€/2

(m – n) x = Ï€ (2k + 1)/2

x = Ï€ (2k + 1)/2(m – n)

âˆ´ the general solution is

x = Ï€ (2k + 1)/2(m – n), where m, n, kÂ ÏµÂ Z.

(ix) tan px = cot qx

Let us simplify,

tan px = cot qx

tan px = tan (Ï€/2 – qx) [since, cot A = tan (Ï€/2 – A)]

px = nÏ€ Â± (Ï€/2 â€“ qx)

(p + q) x = nÏ€ + Ï€/2

x = nÏ€/(p+q) + Ï€/2(p+q)

= Ï€ (2n +1)/ 2(p+q)

âˆ´ the general solution is

x = Ï€ (2n +1)/ 2(p+q), where nÂ ÏµÂ Z.

(x) sin 2x + cos x = 0

Let us simplify,

sin 2x + cos x = 0

cos x = – sin 2x

cos x = – cos (Ï€/2 â€“ 2x) [since, sin A = cos (Ï€/2 – A)]

= cos (Ï€ â€“ (Ï€/2 â€“ 2x)) [since, -cos A = cos (Ï€ – A)]

= cos (Ï€/2 + 2x)

x = 2nÏ€ Â± (Ï€/2 + 2x)

So,

x = 2nÏ€ + (Ï€/2 + 2x) [or] x = 2nÏ€ – (Ï€/2 + 2x)

x = – Ï€/2 – 2nÏ€ [or] 3x = 2nÏ€ – Ï€/2

x = – Ï€/2 (1 + 4n) [or] x = Ï€/6 (4n – 1)

âˆ´ the general solution is

x = – Ï€/2 (1 + 4n), where nÂ ÏµÂ Z. [or] x = Ï€/6 (4n – 1)

x = Ï€/2 (4n – 1), where nÂ ÏµÂ Z. [or] x = Ï€/6 (4n – 1), where nÂ ÏµÂ Z.

(xi) sin x = tan x

Let us simplify,

sin x = tan x

sin x = sin x/cos x

sin x cos x = sin x

sin x (cos x – 1) = 0

So,

Sin x = 0 or cos x â€“ 1 = 0

Sin x = sin 0 [or] cos x = 1

Sin x = sin 0 [or] cos x = cos 0

x = nÏ€ [or] x = 2mÏ€

âˆ´ the general solution is

x = nÏ€ [or] 2mÏ€, where n, mÂ ÏµÂ Z.

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

Let us simplify,

sin 3x + cos 2x = 0

cos 2x = – sin 3x

cos 2x = – cos (Ï€/2 â€“ 3x) [since, sin A = cos (Ï€/2 – A)]

cos 2x = cos (Ï€ â€“ (Ï€/2 â€“ 3x)) [since, -cos A = cos (Ï€ – A)]

cos 2x = cos (Ï€/2 + 3x)

2x = 2nÏ€ Â± (Ï€/2 + 3x)

So,

2x = 2nÏ€ + (Ï€/2 + 3x) [or] 2x = 2nÏ€ – (Ï€/2 + 3x)

x = -Ï€/2 â€“ 2nÏ€ [or] 5x = 2nÏ€ â€“ Ï€/2

x = -Ï€/2 (1 + 4n) [or] x = Ï€/10 (4n – 1)

x = – Ï€/2 (4n + 1) [or] Ï€/10 (4n – 1)

âˆ´ the general solution is

x = – Ï€/2 (4n + 1) [or] Ï€/10 (4n – 1)

x = Ï€/2 (4n – 1) [or] Ï€/10 (4n – 1), where nÂ ÏµÂ Z.

3. Solve the following equations:

(i) sin2 x â€“ cos x = 1/4

(ii) 2 cos2 x â€“ 5 cos x + 2 = 0

(iii) 2 sin2 x + âˆš3 cos x + 1 = 0

(iv) 4 sin2 x â€“ 8 cos x + 1 = 0

(v) tan2 x + (1 – âˆš3) tan x – âˆš3 = 0

(vi) 3 cos2 x – 2âˆš3 sin x cos x â€“ 3 sin2 x = 0

(vii) cos 4x = cos 2x

Solution:

The general solution of any trigonometric equation is given as:

sin x = sin y, implies x = nÏ€ + (â€“ 1)n y, where nÂ âˆˆÂ Z.

cos x = cos y, implies x = 2nÏ€Â Â±Â y, where nÂ âˆˆÂ Z.

tan x = tan y, implies x = nÏ€Â + y, where nÂ âˆˆÂ Z.

(i) sin2 x â€“ cos x = 1/4

Let us simplify,

sin2 x â€“ cos x = Â¼

1 â€“ cos2 x â€“ cos x = 1/4 [since, sin2Â x = 1 â€“ cos2Â x]

4 â€“ 4 cos2 x â€“ 4 cos x = 1

4cos2 x + 4cos x â€“ 3 = 0

Let cos x be â€˜kâ€™

So,

4k2 + 4k â€“ 3 = 0

4k2 â€“ 2k + 6k â€“ 3 = 0

2k (2k – 1) + 3 (2k – 1) = 0

(2k – 1) + (2k + 3) = 0

(2k – 1) = 0 or (2k + 3) = 0

k = 1/2 or k = -3/2

cos x = 1/2 or cos x = -3/2

we shall consider only cos x = 1/2. cos x = -3/2 is not possible.

so,

cos x = cos 60o = cos Ï€/3

x = 2nÏ€ Â± Ï€/3

âˆ´ the general solution is

x = 2nÏ€ Â± Ï€/3, where nÂ ÏµÂ Z.

(ii) 2 cos2 x â€“ 5 cos x + 2 = 0

Let us simplify,

2 cos2 x â€“ 5 cos x + 2 = 0

Let cos x be â€˜kâ€™

2k2Â â€“ 5k + 2 = 0

2k2Â â€“ 4k â€“ k +2 = 0

2k(k â€“ 2) -1(k -2) = 0

(k â€“ 2) (2k – 1) = 0

k = 2 or k = 1/2

cos x = 2 or cos x = 1/2

we shall consider only cos x = 1/2. cos x = 2 is not possible.

so,

cos x = cos 60o = cos Ï€/3

x = 2nÏ€ Â± Ï€/3

âˆ´ the general solution is

x = 2nÏ€ Â± Ï€/3, where nÂ ÏµÂ Z.

(iii) 2 sin2 x + âˆš3 cos x + 1 = 0

Let us simplify,

2 sin2 x + âˆš3 cos x + 1 = 0

2 (1 â€“ cos2 x) + âˆš3 cos x + 1 = 0 [since, sin2Â x = 1 â€“ cos2Â x]

2 â€“ 2 cos2 x + âˆš3 cos x + 1 = 0

2 cos2 x – âˆš3 cos x – 3 = 0

Let cos x be â€˜kâ€™

2k2Â – âˆš3 k â€“ 3 = 0

2k2Â -2âˆš3 k + âˆš3 k â€“ 3 = 0

2k(k â€“ âˆš3) +âˆš3(k â€“ âˆš3) = 0

(2k + âˆš3) (k – âˆš3) = 0

k = âˆš3 or k = -âˆš3/2

cos x = âˆš3 or cos x = -âˆš3/2

we shall consider only cos x = -âˆš3/2. cos x = âˆš3 is not possible.

so,

cos x = -âˆš3/2

cos x = cos 150Â° = cos 5Ï€/6

x = 2nÏ€ Â± 5Ï€/6, where nÂ ÏµÂ Z.

(iv) 4 sin2 x â€“ 8 cos x + 1 = 0

Let us simplify,

4 sin2 x â€“ 8 cos x + 1 = 0

4 (1 â€“ cos2 x) â€“ 8 cos x + 1 = 0 [since, sin2Â x = 1 â€“ cos2Â x]

4 â€“ 4 cos2 x â€“ 8 cos x + 1 = 0

4 cos2 x + 8 cos x – 5 = 0

Let cos x be â€˜kâ€™

4k2Â + 8k â€“ 5 = 0

4k2Â -2k + 10k â€“ 5 = 0

2k(2k â€“ 1) + 5(2k â€“ 1) = 0

(2k + 5) (2k – 1) = 0

k = -5/2 = -2.5 or k = 1/2

cos x = -2.5 or cos x = 1/2

we shall consider only cos x = 1/2. cos x = -2.5 is not possible.

so,

cos x = cos 60o = cos Ï€/3

x = 2nÏ€ Â± Ï€/3

âˆ´ the general solution is

x = 2nÏ€ Â± Ï€/3, where nÂ ÏµÂ Z.

(v) tan2 x + (1 – âˆš3) tan x – âˆš3 = 0

Let us simplify,

tan2 x + (1 – âˆš3) tan x – âˆš3 = 0

tan2 x + tan x – âˆš3 tan x – âˆš3 = 0

tan x (tan x + 1) – âˆš3 (tan x + 1) = 0

(tan x + 1) ( tan x – âˆš3) = 0

tan x = -1 or tan x = âˆš3

As, tan xÂ ÏµÂ (-âˆž , âˆž) so both values are valid and acceptable.

tan x = tan (-Ï€/4) or tan x = tan (Ï€/3)

x = mÏ€ â€“ Ï€/4 or x = nÏ€ + Ï€/3

âˆ´ the general solution is

x = mÏ€ â€“ Ï€/4 or nÏ€ + Ï€/3, where m, nÂ ÏµÂ Z.

(vi) 3 cos2 x – 2âˆš3 sin x cos x â€“ 3 sin2 x = 0

Let us simplify,

3 cos2 x – 2âˆš3 sin x cos x â€“ 3 sin2 x = 0

3 cos2 x – 3âˆš3 sin x cos x + âˆš3 sin x cos x â€“ 3 sin2 x = 0

3 cos x (cos x – âˆš3sin x) + âˆš3 sin x (cos x – âˆš3 sin x) = 0

âˆš3 (cos x – âˆš3 sin x) (âˆš3 cos x + sin x) = 0

cos x – âˆš3 sin x = 0 or sin x + âˆš3 cos x = 0

cos x = âˆš3 sin x or sin x = -âˆš3 cos x

tan x = 1/âˆš3 or tan x = -âˆš3

As, tan xÂ ÏµÂ (-âˆž , âˆž) so both values are valid and acceptable.

tan x = tan (Ï€/6) or tan x = tan (-Ï€/3)

x = mÏ€ + Ï€/6 or x = nÏ€ â€“ Ï€/3

âˆ´ the general solution is

x = mÏ€ + Ï€/6 or nÏ€ â€“ Ï€/3, where m, nÂ ÏµÂ Z.

(vii) cos 4x = cos 2x

Let us simplify,

cos 4x = cos 2x

4x = 2nÏ€ Â± 2x

So,

4x = 2nÏ€ + 2x [or] 4x = 2nÏ€ – 2x

2x = 2nÏ€ [or] 6x = 2nÏ€

x = nÏ€ [or] x = nÏ€/3

âˆ´ the general solution is

x = nÏ€ [or] nÏ€/3, where nÂ ÏµÂ Z.

4. Solve the following equations:

(i) cos x + cos 2x + cos 3x = 0

(ii) cos x + cos 3x â€“ cos 2x = 0

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

(iv) cos x cos 2x cos 3x = 1/4

(v) cos x + sin x = cos 2x + sin 2x

(vi) sin x + sin 2x + sin 3x = 0

(vii) sin x + sin 2x + sin 3x + sin 4x = 0

(viii) sin 3x â€“ sin x = 4 cos2 x – 2

(ix) sin 2x â€“ sin 4x + sin 6x = 0

Solution:

The general solution of any trigonometric equation is given as:

sin x = sin y, implies x = nÏ€ + (â€“ 1)n y, where nÂ âˆˆÂ Z.

cos x = cos y, implies x = 2nÏ€Â Â±Â y, where nÂ âˆˆÂ Z.

tan x = tan y, implies x = nÏ€Â + y, where nÂ âˆˆÂ Z.

(i) cos x + cos 2x + cos 3x = 0

Let us simplify,

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

we shall rearrange and use transformation formula

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

by using the formula, cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2

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

cos 2x + 2cos 2x cos x = 0

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

cos 2x = 0 or 1 + 2cos x = 0

cos 2x = cos 0 or cos x = -1/2

cos 2x = cos Ï€/2 or cos x = cos (Ï€ – Ï€/3)

cos 2x = cos Ï€/2 or cos x = cos (2Ï€/3)

2x = (2n + 1) Ï€/2 or x = 2mÏ€ Â± 2Ï€/3

x = (2n + 1) Ï€/4 or x = 2mÏ€ Â± 2Ï€/3

âˆ´ the general solution is

x = (2n + 1) Ï€/4 or 2mÏ€ Â± 2Ï€/3, where m, nÂ ÏµÂ Z.

(ii) cos x + cos 3x â€“ cos 2x = 0

Let us simplify,

cos x + cos 3x â€“ cos 2x = 0

we shall rearrange and use transformation formula

cos x â€“ cos 2x + cos 3x = 0

– cos 2x + (cos x + cos 3x) = 0

By using the formula, cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2

– cos 2x + 2 cos (3x+x)/2 cos (3x-x)/2 = 0

– cos 2x + 2cos 2x cos x = 0

cos 2x ( -1 + 2 cos x) = 0

cos 2x = 0 or -1 + 2cos x = 0

cos 2x = cos 0 or cos x = 1/2

cos 2x = cos Ï€/2 or cos x = cos (Ï€/3)

2x = (2n + 1) Ï€/2 or x = 2mÏ€ Â± Ï€/3

x = (2n + 1) Ï€/4 or x = 2mÏ€ Â± Ï€/3

âˆ´ the general solution is

x = (2n + 1) Ï€/4 or 2mÏ€ Â± Ï€/3, where m, nÂ ÏµÂ Z.

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

Let us simplify,

sin x + sin 5x = sin 3x

sin x + sin 5x – sin 3x = 0

we shall rearrange and use transformation formula

– sin 3x + sin x + sin 5x = 0

– sin 3x + (sin 5x + sin x) = 0

By using the formula, sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

– sin 3x + 2 sin (5x+x)/2 cos (5x-x)/2 = 0

2sin 3x cos 2x â€“ sin 3x = 0

sin 3x ( 2cos 2x â€“ 1) = 0

sin 3x = 0 or 2cos 2x â€“ 1 = 0

sin 3x = sin 0 or cos 2x = 1/2

sin 3x = sin 0 or cos 2x = cos Ï€/3

3x = nÏ€ or 2x = 2mÏ€ Â± Ï€/3

x = nÏ€/3 or x = mÏ€ Â± Ï€/6

âˆ´ the general solution is

x = nÏ€/3 or mÏ€ Â± Ï€/6, where m, nÂ ÏµÂ Z.

(iv) cos x cos 2x cos 3x = 1/4

Let us simplify,

cos x cos 2x cos 3x = 1/4

4 cos x cos 2x cos 3x â€“ 1 = 0

By using the formula,

2 cos A cos B = cos (A + B) + cos (A â€“ B)

2(2cos x cos 3x) cos 2x â€“ 1 = 0

2(cos 4x + cos 2x) cos2x â€“ 1 = 0

2(2cos2Â 2x â€“ 1 + cos 2x) cos 2x â€“ 1 = 0 [using cos 2A = 2cos2A â€“ 1]

4cos3Â 2x â€“ 2cos 2x + 2cos2Â 2x â€“ 1 = 0

2cos2Â 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

(2cos2Â 2x â€“ 1) (2 cos 2x + 1) = 0

So,

2cos 2x + 1 = 0 or (2cos2Â 2x â€“ 1) = 0

cos 2x = -1/2 or cos 4x = 0 [using cos 2Î¸ = 2cos2Î¸ â€“ 1]

cos 2x = cos (Ï€ – Ï€/3) or cos 4x = cos Ï€/2

cos 2x = cos 2Ï€/3 or cos 4x = cos Ï€/2

2x = 2mÏ€ Â± 2Ï€/3 or 4x = (2n + 1) Ï€/2

x = mÏ€ Â± Ï€/3 or x = (2n + 1) Ï€/8

âˆ´ the general solution is

x = mÏ€ Â± Ï€/3 or (2n + 1) Ï€/8, where m, nÂ ÏµÂ Z.

(v) cos x + sin x = cos 2x + sin 2x

Let us simplify,

cos x + sin x = cos 2x + sin 2x

upon rearranging we get,

cos x â€“ cos 2x = sin 2x â€“ sin x

By using the formula,

sin A â€“ sin B = 2 cos (A+B)/2 sin (A-B)/2

cos A â€“ cos B = – 2 sin (A+B)/2 sin (A-B)/2

So,

-2 sin (2x+x)/2 sin (2x-x)/2 = 2 cos (2x+x)/2 sin (2x-x)/2

2 sin 3x/2 sin x/2 = 2 cos 3x/2 sin x/2

Sin x/2 (sin 3x/2 â€“ cos 3x/2) = 0

So,

Sin x/2 = 0 or sin 3x/2 = cos 3x/2

Sin x/2 = sin mÏ€ or sin 3x/2 / cos 3x/2 = 0

Sin x/2 = sin mÏ€ or tan 3x/2 = 1

Sin x/2 = sin mÏ€ or tan 3x/2 = tan Ï€/4

x/2 = mÏ€ or 3x/2 = nÏ€ + Ï€/4

x = 2mÏ€ or x = 2nÏ€/3 + Ï€/6

âˆ´ the general solution is

x = 2mÏ€ or 2nÏ€/3 + Ï€/6, where m, nÂ ÏµÂ Z.

(vi) sin x + sin 2x + sin 3x = 0

Let us simplify,

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

we shall rearrange and use transformation formula

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

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

So,

Sin 2x + 2 sin (3x+x)/2 cos (3x-x)/2 = 0

Sin 2x + 2sin 2x cos x = 0

Sin 2x (2 cos x + 1) = 0

Sin 2x = 0 or 2cos x + 1 = 0

Sin 2x = sin 0 or cos x = -1/2

Sin 2x = sin 0 or cos x = cos (Ï€ â€“ Ï€/3)

Sin 2x = sin 0 or cos x = cos 2Ï€/3

2x = nÏ€ or x = 2mÏ€ Â± 2Ï€/3

x = nÏ€/2 or x = 2mÏ€ Â± 2Ï€/3

âˆ´ the general solution is

x = nÏ€/2 or 2mÏ€ Â± 2Ï€/3, where m, nÂ ÏµÂ Z.

(vii) sin x + sin 2x + sin 3x + sin 4x = 0

Let us simplify,

sin x + sin 2x + sin 3x + sin 4x = 0

we shall rearrange and use transformation formula

sin x + sin 3x + sin 2x + sin 4x = 0

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

So,

2 sin (3x+x)/2 cos (3x-x)/2 + 2 sin (4x+2x)/2 cos (4x-2x)/2 = 0

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

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

Again by using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

we get,

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

2cos x (2 sin 5x/2 cos x/2) = 0

4 cos x sin 5x/2 cos x/2 = 0

So,

Cos x = 0 or sin 5x/2 = 0 or cos x/2 = 0

Cos x = cos 0 or sin 5x/2 = sin 0 or cos x/2 = cos 0

Cos x = cos Ï€/2 or sin 5x/2 = kÏ€ or cos x/2 = cos (2p + 1) Ï€/2

x = (2n + 1) Ï€/2 or 5x/2 = kÏ€ or x/2 = (2p + 1) Ï€/2

x = (2n + 1) Ï€/2 or x = 2kÏ€/5 or x = (2p + 1)

x = nÏ€ + Ï€/2 or x = 2kÏ€/5 or x = (2p + 1)

âˆ´ the general solution is

x = nÏ€ + Ï€/2 or x = 2kÏ€/5 or x = (2p + 1), where n, k, pÂ ÏµÂ Z.

(viii) sin 3x â€“ sin x = 4 cos2 x – 2

Let us simplify,

sin 3x â€“ sin x = 4 cos2 x – 2

sin 3x â€“ sin x = 2(2 cos2Â x â€“ 1)

sin 3x â€“ sin x = 2 cos 2x [since,Â cos 2A = 2cos2Â A â€“ 1]

By using the formula,

Sin A â€“ sin B = 2 cos (A+B)/2 sin (A-B)/2

So,

2 cos (3x+x)/2 sin (3x-x)/2 = 2 cos 2x

2 cos 2x sin x â€“ 2 cos 2x = 0

2 cos 2x (sin x – 1) = 0

Then,

2 cos 2x = 0 or sin x â€“ 1 = 0

Cos 2x = 0 or sin x = 1

Cos 2x = cos 0 or sin x = sin 1

Cos 2x = cos 0 or sin x = sin Ï€/2

2x = (2n + 1) Ï€/2 or x = mÏ€ + (-1) m Ï€/2

x = (2n + 1) Ï€/4 or x = mÏ€ + (-1) m Ï€/2

âˆ´ the general solution is

x = (2n + 1) Ï€/4 or mÏ€ + (-1) m Ï€/2, where m, nÂ ÏµÂ Z.

(ix) sin 2x â€“ sin 4x + sin 6x = 0

Let us simplify,

sin 2x â€“ sin 4x + sin 6x = 0

we shall rearrange and use transformation formula

– sin 4x + sin 6x + sin 2x = 0

By using the formula,

sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2

we get,

– sin 4x + 2 sin (6x+2x)/2 cos (6x-2x)/2 = 0

– sin 4x + 2 sin 4x cos 2x = 0

Sin 4x (2 cos 2x – 1) = 0

So,

Sin 4x = 0 or 2 cos 2x â€“ 1 = 0

Sin 4x = sin 0 or cos 2x = 1/2

Sin 4x = sin 0 or cos 2x = Ï€/3

4x = nÏ€ or 2x = 2mÏ€ Â± Ï€/3

x = nÏ€/4 or x = mÏ€ Â± Ï€/6

âˆ´ the general solution is

x = nÏ€/4 or mÏ€ Â± Ï€/6, where m, nÂ ÏµÂ Z.

5. Solve the following equations:

(i) tan x + tan 2x + tan 3x = 0

(ii) tan x + tan 2x = tan 3x

(iii) tan 3x + tan x = 2 tan 2x

Solution:

The general solution of any trigonometric equation is given as:

sin x = sin y, implies x = nÏ€ + (â€“ 1)n y, where nÂ âˆˆÂ Z.

cos x = cos y, implies x = 2nÏ€Â Â±Â y, where nÂ âˆˆÂ Z.

tan x = tan y, implies x = nÏ€Â + y, where nÂ âˆˆÂ Z.

(i) tan x + tan 2x + tan 3x = 0

Let us simplify,

tan x + tan 2x + tan 3x = 0

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

By using the formula,

tan (A+B) = [tan A + tan B] / [1 â€“ tan A tan B]

So,

tan x + tan 2x + [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

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

(tan x + tan 2x) ([2 â€“ tan x tan 2x] / [1 â€“ tan x tan 2x]) = 0

Then,

(tan x + tan 2x) = 0 or ([2 â€“ tan x tan 2x] / [1 â€“ tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [2 â€“ tan x tan 2x] = 0

tan x = tan (-2x) or tan x tan 2x = 2

x = nÏ€ + (-2x) or tax x [2tan x/(1 â€“ tan2 x)] = 2 [Using, tan 2x = 2 tan x / 1-tan2 x]

3x = nÏ€ or 2 tan2 x / (1-tan2 x) = 2

3x = nÏ€ or 2 tan2 x = 2(1 â€“ tan2 x)

3x = nÏ€ or 2 tan2 x = 2 – 2tan2 x

3x = nÏ€ or 4 tan2 x = 2

x = nÏ€/3 or tan2 x = 2/4

x = nÏ€/3 or tan2 x = 1/2

x = nÏ€/3 or tan x = 1/âˆš2

x = nÏ€/3 or x = tan Î± [let 1/âˆš2 be â€˜Î±â€™]

x = nÏ€/3 or x = mÏ€ + Î±

âˆ´ the general solution is

x = nÏ€/3 or mÏ€ + Î±, where Î± = tan-11/âˆš2, m, n âˆˆÂ Z.

(ii) tan x + tan 2x = tan 3x

Let us simplify,

tan x + tan 2x = tan 3x

tan x + tan 2x â€“ tan 3x = 0

tan x + tan 2x â€“ tan (x + 2x) = 0

By using the formula,

tan (A+B) = [tan A + tan B] / [1 â€“ tan A tan B]

So,

tan x + tan 2x – [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

(tan x + tan 2x) (1 – 1/(1- tan x tan 2x)) = 0

(tan x + tan 2x) ([â€“ tan x tan 2x] / [1 â€“ tan x tan 2x]) = 0

Then,

(tan x + tan 2x) = 0 or ([â€“ tan x tan 2x] / [1 â€“ tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [â€“ tan x tan 2x] = 0

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

tan x = tan (-2x) or 2tan2 x / (1 â€“ tan2 x) = 0 [Using, tan 2x = 2 tan x / 1-tan2 x]

x = nÏ€ + (-2x) or x = mÏ€ + 0

3x = nÏ€ or x = mÏ€

x = nÏ€/3 or x = mÏ€

âˆ´ the general solution is

x = nÏ€/3 or mÏ€, where m, n âˆˆÂ Z.

(iii) tan 3x + tan x = 2 tan 2x

Let us simplify,

tan 3x + tan x = 2 tan 2x

tan 3x + tan x = tan 2x + tan 2x

upon rearranging we get,

tan 3x – tan 2x = tan 2x – tan x

By using the formula,

tan (A-B) = [tan A â€“ tan B] / [1 + tan A tan B]

so,

[(tan 3x â€“ tan 2x) (1+tan 3x tan 2x)] / [1 + tan 3x tan 2x] = [(tan 2x-tan x) (1+tan x tan 2x)] / [1 + tan 2x tan x]

tan (3x â€“ 2x) (1 + tan 3x tan 2x) = tan (2x – x) (1 + tan x tan 2x)

tan x [1 + tan 3x tan 2x â€“ 1 â€“ tan 2x tan x] = 0

tan x tan 2x (tan 3x â€“ tan x) = 0

so,

tan x = 0 or tan 2x = 0 or (tan 3x â€“ tan x) = 0

tan x = 0 or tan 2x = 0 or tan 3x = tan x

x = nÏ€ or 2x = mÏ€ or 3x = kÏ€ + x

x = nÏ€ or x = mÏ€/2 or 2x = kÏ€

x = nÏ€ or x = mÏ€/2 or x = kÏ€/2

âˆ´ the general solution is

x = nÏ€ or mÏ€/2 or kÏ€/2, where, m, n, k âˆˆÂ Z.

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 11

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