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# RD Chapter 11- Trigonometric Equations Interview Questions Answers

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Question 1 :
Find the general solutions of the following equations:
(i) sin x = 1/2
(ii) cos x = – √3/2
(iii) cosec x = – √2
(iv) sec x = √2
(v) tan x = -1/√3
(vi) √3 sec x = 2

The general solution of any trigonometric equation is givenas:

sin x = sin y, implies x = nπ + (– 1)y,where n  Z.

cos x = cos y, implies x = 2nπ ± y, where n  Z.

tan x = tan y, implies x = nπ + y, where n  Z.

(i) sin x = 1/2

We know sin 30o = sin π/6 = ½

So,

Sin x = sin π/6

the general solution is

x = nπ + (– 1) n π/6, where n  Z. [since, sin x = sin A => x = nπ + (– 1) n A]

(ii) cos x = – √3/2

We know, cos 150o = (- √3/2) = cos 5π/6

So,

Cos x = cos 5π/6

the general solution is

x = 2nπ ± 5π/6, where n ϵ Z.

(iii) cosec x = – √2

Let us simplify,

1/sin x = – √2 [since, cosec x = 1/sin x]

Sin x = -1/√2

= sin [π + π/4]

= sin 5π/4 or sin (-π/4)

the general solution is

x = nπ + (-1)n+1 π/4, where n ϵ Z.

(iv) sec x = √2

Let us simplify,

1/cos x = √2 [since, sec x = 1/cos x]

Cos x = 1/√2

= cos π/4

the general solution is

x = 2nπ ± π/4, where n ϵ Z.

(v) tan x = -1/√3

Let us simplify,

tan x = -1/√3

tan x = tan (π/6)

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

the general solution is

x = nπ + (-π/6), where n ϵ Z.

or x = nπ – π/6, where n ϵ Z.

(vi) √3 sec x = 2

Let us simplify,

sec x = 2/√3

1/cos x = 2/√3

Cos x = √3/2

= cos (π/6)

the general solution is

x = 2nπ ± π/6, where n ϵ Z.

Question 2 :
Find the general solutions of the following equations:
(i) sin 2x = √3/2
(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

The general solution of any trigonometric equation is givenas:

sin x = sin y, implies x = nπ + (– 1)y,where n  Z.

cos x = cos y, implies x = 2nπ ± y, where n  Z.

tan x = tan y, implies x = nπ + y, where n  Z.

(i) sin 2x = √3/2

Let us simplify,

sin 2x = √3/2

= sin (π/3)

the general solution is

2x = nπ + (-1)n π/3, where n ϵ Z.

x = nπ/2 + (-1)n π/6, where n ϵ Z.

(ii) cos 3x = 1/2

Let us simplify,

cos 3x = 1/2

= cos (π/3)

the general solution is

3x = 2nπ ± π/3, where n ϵ Z.

x = 2nπ/3 ± π/9, where n ϵ Z.

(iii) sin 9x = sin x

Let us simplify,

Sin 9x – sin x = 0

Using transformation formula,

Sin A – 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 π/2

5x = (2n + 1)π/2

x = (2n + 1)π/10, where n ϵ Z.

sin 4x = 0

sin 4x = sin 0

4x = nπ

x = nπ/4, where n ϵ Z.

the general solution is

x = (2n + 1)π/10 or nπ/4, where n ϵ Z.

(iv) sin 2x = cos 3x

Let us simplify,

sin 2x = cos 3x

cos (π/2 – 2x) = cos 3x [since, sin A = cos (π/2 – A)]

π/2 – 2x = 2nπ ± 3x

π/2 – 2x = 2nπ + 3x [or] π/2 – 2x = 2nπ – 3x

5x = π/2 + 2nπ [or] x = 2nπ – π/2

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

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

the general solution is

x = π/10 (4n + 1) [or] x = π/2 (4n – 1), wheren ϵ Z.

(v) tan x + cot 2x = 0

Let us simplify,

tan x = – cot 2x

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

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

x = nπ + 2x – π/2

x = nπ – π/2

the general solution is

x = nπ – π/2, where n ϵ Z.

(vi) tan 3x = cot x

Let us simplify,

tan 3x = cot x

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

3x = nπ + π/2 – x

4x = nπ + π/2

x = nπ/4 + π/8

the general solution is

x = nπ/4 + π/8, where n ϵ 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 (π/2 – x) [since, cot A = tan (π/2 – A)]

2x = nπ + π/2 – x

3x = nπ + π/2

x = nπ/3 + π/6

the general solution is

x = nπ/3 + π/6, where n ϵ Z.

(viii) tan mx + cot nx = 0

Let us simplify,

tan mx + cot nx = 0

tan mx = – cot nx

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

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

mx = kπ + nx + π/2

(m – n) x = kπ + π/2

(m – n) x = π (2k + 1)/2

x = π (2k + 1)/2(m – n)

the general solution is

x = π (2k + 1)/2(m – n), where m, n, k ϵ Z.

(ix) tan px = cot qx

Let us simplify,

tan px = cot qx

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

px = nπ ± (π/2 – qx)

(p + q) x = nπ + π/2

x = nπ/(p+q) + π/2(p+q)

= π (2n +1)/ 2(p+q)

the general solution is

x = π (2n +1)/ 2(p+q), where n ϵ Z.

(x) sin 2x + cos x = 0

Let us simplify,

sin 2x + cos x = 0

cos x = – sin 2x

cos x = – cos (π/2 – 2x) [since, sin A = cos (π/2 – A)]

= cos (π – (π/2 – 2x)) [since, -cos A = cos (π – A)]

= cos (π/2 + 2x)

x = 2nπ ± (π/2 + 2x)

So,

x = 2nπ + (π/2 + 2x) [or] x = 2nπ – (π/2 + 2x)

x = – π/2 – 2nπ [or] 3x = 2nπ – π/2

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

the general solution is

x = – π/2 (1 + 4n), where n ϵ Z. [or] x = π/6 (4n– 1)

x = π/2 (4n – 1), where n ϵ Z. [or] x = π/6 (4n –1), where n ϵ 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 – 1 = 0

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

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

x = nπ [or] x = 2mπ

the general solution is

x = nπ [or] 2mπ, where n, m ϵ Z.

(xii) sin 3x + cos 2x = 0

Let us simplify,

sin 3x + cos 2x = 0

cos 2x = – sin 3x

cos 2x = – cos (π/2 – 3x) [since, sin A = cos (π/2 – A)]

cos 2x = cos (π – (π/2 – 3x)) [since, -cos A = cos (π – A)]

cos 2x = cos (π/2 + 3x)

2x = 2nπ ± (π/2 + 3x)

So,

2x = 2nπ + (π/2 + 3x) [or] 2x = 2nπ – (π/2 + 3x)

x = -π/2 – 2nπ [or] 5x = 2nπ – π/2

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

x = – π/2 (4n + 1) [or] π/10 (4n – 1)

the general solution is

x = – π/2 (4n + 1) [or] π/10 (4n – 1)

x = π/2 (4n – 1) [or] π/10 (4n – 1), where n ϵ Z.

Question 3 :

Solve the following equations:

(i) sin2 x – cos x= 1/4

(ii) 2 cos2 x – 5cos x + 2 = 0

(iii) 2 sin2 x +√3 cos x + 1 = 0

(iv) 4 sin2 x – 8cos x + 1 = 0

(v) tan2 x + (1 –√3) tan x – √3 = 0

(vi) 3 cos2 x –2√3 sin x cos x – 3 sin2 x = 0

(vii) cos 4x = cos 2x

The general solution of any trigonometric equation is givenas:

sin x = sin y, implies x = nπ + (– 1)y,where n  Z.

cos x = cos y, implies x = 2nπ ± y, where n  Z.

tan x = tan y, implies x = nπ + y, where n  Z.

(i) sin2 x – cos x = 1/4

Let us simplify,

sin2 x – cos x = ¼

1 – cos2 x – cos x = 1/4 [since, sin2 x= 1 – cos2 x]

4 – 4 cos2 x – 4 cos x = 1

4cos2 x + 4cos x – 3 = 0

Let cos x be ‘k’

So,

4k2 + 4k – 3 = 0

4k2 – 2k + 6k – 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 notpossible.

so,

cos x = cos 60o = cos π/3

x = 2nπ ± π/3

the general solution is

x = 2nπ ± π/3, where n ϵ Z.

(ii) 2 cos2 x – 5 cos x + 2 = 0

Let us simplify,

2 cos2 x – 5 cos x + 2 = 0

Let cos x be ‘k’

2k2 – 5k + 2 = 0

2k2 – 4k – k +2 = 0

2k(k – 2) -1(k -2) = 0

(k – 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 notpossible.

so,

cos x = cos 60o = cos π/3

x = 2nπ ± π/3

the general solution is

x = 2nπ ± π/3, where n ϵ Z.

(iii) 2 sin2 x + √3 cos x + 1 = 0

Let us simplify,

2 sin2 x + √3 cos x + 1 = 0

2 (1 – cos2 x) + √3 cos x + 1 = 0 [since,sin2 x = 1 – cos2 x]

2 – 2 cos2 x + √3 cos x + 1 = 0

2 cos2 x – √3 cos x – 3 = 0

Let cos x be ‘k’

2k2 – √3 k – 3 = 0

2k2 -2√3 k + √3 k – 3 = 0

2k(k – √3) +√3(k – √3) = 0

(2k + √3) (k – √3) = 0

k = √3 or k = -√3/2

cos x = √3 or cos x = -√3/2

we shall consider only cos x = -√3/2. cos x = √3 is notpossible.

so,

cos x = -√3/2

cos x = cos 150° = cos 5π/6

x = 2nπ ± 5π/6, where n ϵ Z.

(iv) 4 sin2 x – 8 cos x + 1 = 0

Let us simplify,

4 sin2 x – 8 cos x + 1 = 0

4 (1 – cos2 x) – 8 cos x + 1 = 0 [since, sin2 x= 1 – cos2 x]

4 – 4 cos2 x – 8 cos x + 1 = 0

4 cos2 x + 8 cos x – 5 = 0

Let cos x be ‘k’

4k2 + 8k – 5 = 0

4k2 -2k + 10k – 5 = 0

2k(2k – 1) + 5(2k – 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 notpossible.

so,

cos x = cos 60o = cos π/3

x = 2nπ ± π/3

the general solution is

x = 2nπ ± π/3, where n ϵ Z.

(v) tan2 x + (1 – √3) tan x – √3 = 0

Let us simplify,

tan2 x + (1 – √3) tan x – √3 = 0

tan2 x + tan x – √3 tan x – √3 = 0

tan x (tan x + 1) – √3 (tan x + 1) = 0

(tan x + 1) ( tan x – √3) = 0

tan x = -1 or tan x = √3

As, tan x ϵ (-∞ , ∞) so both values are valid andacceptable.

tan x = tan (-π/4) or tan x = tan (π/3)

x = mπ – π/4 or x = nπ + π/3

the general solution is

x = mπ – π/4 or nπ + π/3, where m, n ϵ Z.

(vi) 3 cos2 x – 2√3 sin x cos x – 3 sin2 x= 0

Let us simplify,

3 cos2 x – 2√3 sin x cos x – 3 sin2 x= 0

3 cos2 x – 3√3 sin x cos x + √3 sin x cos x– 3 sin2 x = 0

3 cos x (cos x – √3sin x) + √3 sin x (cos x – √3 sin x) = 0

√3 (cos x – √3 sin x) (√3 cos x + sin x) = 0

cos x – √3 sin x = 0 or sin x + √3 cos x = 0

cos x = √3 sin x or sin x = -√3 cos x

tan x = 1/√3 or tan x = -√3

As, tan x ϵ (-∞ , ∞) so both values are valid andacceptable.

tan x = tan (π/6) or tan x = tan (-π/3)

x = mπ + π/6 or x = nπ – π/3

the general solution is

x = mπ + π/6 or nπ – π/3, where m, n ϵ Z.

(vii) cos 4x = cos 2x

Let us simplify,

cos 4x = cos 2x

4x = 2nπ ± 2x

So,

4x = 2nπ + 2x [or] 4x = 2nπ – 2x

2x = 2nπ [or] 6x = 2nπ

x = nπ [or] x = nπ/3

the general solution is

x = nπ [or] nπ/3, where n ϵ Z.

Question 4 :

Solve the following equations:

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

(ii) cos x + cos 3x – 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 + sin2x

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

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

(viii) sin 3x – sin x = 4 cos2 x– 2

(ix) sin 2x – sin 4x + sin 6x = 0

The general solution of any trigonometric equation is givenas:

sin x = sin y, implies x = nπ + (– 1)y,where n  Z.

cos x = cos y, implies x = 2nπ ± y, where n  Z.

tan x = tan y, implies x = nπ + y, where n  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 π/2 or cos x = cos (π – π/3)

cos 2x = cos π/2 or cos x = cos (2π/3)

2x = (2n + 1) π/2 or x = 2mπ ± 2π/3

x = (2n + 1) π/4 or x = 2mπ ± 2π/3

the general solution is

x = (2n + 1) π/4 or 2mπ ± 2π/3, where m, n ϵ Z.

(ii) cos x + cos 3x – cos 2x = 0

Let us simplify,

cos x + cos 3x – cos 2x = 0

we shall rearrange and use transformation formula

cos x – 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 π/2 or cos x = cos (π/3)

2x = (2n + 1) π/2 or x = 2mπ ± π/3

x = (2n + 1) π/4 or x = 2mπ ± π/3

the general solution is

x = (2n + 1) π/4 or 2mπ ± π/3, where m, n ϵ 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 – sin 3x = 0

sin 3x ( 2cos 2x – 1) = 0

sin 3x = 0 or 2cos 2x – 1 = 0

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

sin 3x = sin 0 or cos 2x = cos π/3

3x = nπ or 2x = 2mπ ± π/3

x = nπ/3 or x = mπ ± π/6

the general solution is

x = nπ/3 or mπ ± π/6, where m, n ϵ 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 – 1 = 0

By using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2(2cos x cos 3x) cos 2x – 1 = 0

2(cos 4x + cos 2x) cos2x – 1 = 0

2(2cos2 2x – 1 + cos 2x) cos 2x – 1 = 0[using cos 2A = 2cos2A – 1]

4cos3 2x – 2cos 2x + 2cos2 2x– 1 = 0

2cos2 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

(2cos2 2x – 1) (2 cos 2x + 1) = 0

So,

2cos 2x + 1 = 0 or (2cos2 2x – 1) = 0

cos 2x = -1/2 or cos 4x = 0 [using cos 2θ = 2cos2θ– 1]

cos 2x = cos (π – π/3) or cos 4x = cos π/2

cos 2x = cos 2π/3 or cos 4x = cos π/2

2x = 2mπ ± 2π/3 or 4x = (2n + 1) π/2

x = mπ ± π/3 or x = (2n + 1) π/8

the general solution is

x = mπ ± π/3 or (2n + 1) π/8, where m, n ϵ 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 – cos 2x = sin 2x – sin x

By using the formula,

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

cos A – 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 – cos 3x/2) = 0

So,

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

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

Sin x/2 = sin mπ or tan 3x/2 = 1

Sin x/2 = sin mπ or tan 3x/2 = tan π/4

x/2 = mπ or 3x/2 = nπ + π/4

x = 2mπ or x = 2nπ/3 + π/6

the general solution is

x = 2mπ or 2nπ/3 + π/6, where m, n ϵ 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 (π – π/3)

Sin 2x = sin 0 or cos x = cos 2π/3

2x = nπ or x = 2mπ ± 2π/3

x = nπ/2 or x = 2mπ ± 2π/3

the general solution is

x = nπ/2 or 2mπ ± 2π/3, where m, n ϵ 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 π/2 or sin 5x/2 = kπ or cos x/2 = cos (2p + 1)π/2

x = (2n + 1) π/2 or 5x/2 = kπ or x/2 = (2p + 1) π/2

x = (2n + 1) π/2 or x = 2kπ/5 or x = (2p + 1)

x = nπ + π/2 or x = 2kπ/5 or x = (2p + 1)

the general solution is

x = nπ + π/2 or x = 2kπ/5 or x = (2p + 1), where n, k,p ϵ Z.

(viii) sin 3x – sin x = 4 cos2 x – 2

Let us simplify,

sin 3x – sin x = 4 cos2 x – 2

sin 3x – sin x = 2(2 cos2 x – 1)

sin 3x – sin x = 2 cos 2x [since, cos 2A = 2cos2 A– 1]

By using the formula,

Sin A – 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 – 2 cos 2x = 0

2 cos 2x (sin x – 1) = 0

Then,

2 cos 2x = 0 or sin x – 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 π/2

2x = (2n + 1) π/2 or x = mπ + (-1) m π/2

x = (2n + 1) π/4 or x = mπ + (-1) m π/2

the general solution is

x = (2n + 1) π/4 or mπ + (-1) m π/2,where m, n ϵ Z.

(ix) sin 2x – sin 4x + sin 6x = 0

Let us simplify,

sin 2x – 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 – 1 = 0

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

Sin 4x = sin 0 or cos 2x = π/3

4x = nπ or 2x = 2mπ ± π/3

x = nπ/4 or x = mπ ± π/6

the general solution is

x = nπ/4 or mπ ± π/6, where m, n ϵ Z.

Question 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

The general solution of any trigonometric equation is givenas:

sin x = sin y, implies x = nπ + (– 1)y,where n  Z.

cos x = cos y, implies x = 2nπ ± y, where n  Z.

tan x = tan y, implies x = nπ + y, where n  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 – 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 – tan x tan 2x] / [1 – tan x tan 2x]) =0

Then,

(tan x + tan 2x) = 0 or ([2 – tan x tan 2x] / [1 – tan x tan2x]) = 0

(tan x + tan 2x) = 0 or [2 – tan x tan 2x] = 0

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

x = nπ + (-2x) or tax x [2tan x/(1 – tan2 x)]= 2 [Using, tan 2x = 2 tan x / 1-tan2 x]

3x = nπ or 2 tan2 x / (1-tan2 x)= 2

3x = nπ or 2 tan2 x = 2(1 – tan2 x)

3x = nπ or 2 tan2 x = 2 – 2tan2 x

3x = nπ or 4 tan2 x = 2

x = nπ/3 or tan2 x = 2/4

x = nπ/3 or tan2 x = 1/2

x = nπ/3 or tan x = 1/√2

x = nπ/3 or x = tan α [let 1/√2 be ‘α’]

x = nπ/3 or x = mπ + α

the general solution is

x = nπ/3 or mπ + α, where α = tan-11/√2, m, n  Z.

(ii) tan x + tan 2x = tan 3x

Let us simplify,

tan x + tan 2x = tan 3x

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 – 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) ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0

Then,

(tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan2x]) = 0

(tan x + tan 2x) = 0 or [– tan x tan 2x] = 0

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

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

x = nπ + (-2x) or x = mπ + 0

3x = nπ or x = mπ

x = nπ/3 or x = mπ

the general solution is

x = nπ/3 or mπ, where m, n  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 – tan B] / [1 + tan A tan B]

so,

[(tan 3x – tan 2x) (1+tan 3x tan2x)] / [1 + tan 3x tan 2x] = [(tan 2x-tan x) (1+tan x tan 2x)] / [1 + tan 2xtan x]

tan (3x – 2x) (1 + tan 3x tan 2x) = tan (2x – x) (1 + tan xtan 2x)

tan x [1 + tan 3x tan 2x – 1 – tan 2x tan x] = 0

tan x tan 2x (tan 3x – tan x) = 0

so,

tan x = 0 or tan 2x = 0 or (tan 3x – tan x) = 0

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

x = nπ or 2x = mπ or 3x = kπ + x

x = nπ or x = mπ/2 or 2x = kπ

x = nπ or x = mπ/2 or x = kπ/2

the general solution is

x = nπ or mπ/2 or kπ/2, where, m, n, k  Z.

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