4 Trigonometric Equations
GENERAL SOLUTIONS OF T.E. θ + α
Or cos = 0 ⇒
θ+ α
= (2n + 1)
π
2 2 2
Types of Solutions: θ = (2n + 1)π – α
General solution: All angles π – α, 3π – α, 5π – α, – π – α ...
Principle solution: θ Є [0, 2π)
Particular solution: All solutions in the given θ = nπ + (–1)nα, n Є I
interval.
Note Proof:
sin x = 0 ↔ x = nπ cos θ = cos α ⇒ cos θ – cos α = 0
π θ + α θ − α
cos x = 0 ↔ x = (2n + 1)
2 −2 sin sin = 0
2 2
π
sin x = 1 ↔ x = (4n + 1) θ = 2nπ – α (or) θ = 2nπ + α
2
π ⇒ θ = 2nπ ± α
sin x = –1 ↔ x = (4n – 1)
2
cos x = 1 ↔ x = 2nπ Question
cos x = –1 ↔ x = (2n – 1)π
(Where 'n' is any integer) Q. Solve following equation: tan 3θ = –1
π
Sol. Sumit: tan 3θ = tan −
ONLY FOUR FORMULA'S 4
π π(4n − 1)
3θ = nπ − =
4 4
(1) sin θ = sin α ⇒ θ = nπ + (–1)nα, (n Є I)
π
(2) cos θ = cos α ⇒ θ = 2nπ ± α, (n Є I) θ = (4n − 1)
12 Same
(3) tan θ = tan α ⇒ θ = nπ + α, (n Є I) 3π
(4) sin2 θ = sin2 α Santosh: tan 3θ = tan
4
cos2 θ = cos2 α ⇒ θ = nπ ± α, (n Є I) 3π π
3θ = nπ + ⇒ θ = (4n + 3)
tan2 θ = tan2 α 4 12
Question
Proof:
Q. Solve following equation: cos 3θ = cos θ
sin θ = sin α ⇒ sin θ – sin α = 0
Sol. cos 3θ = cos θ
θ − α θ + α
2 sin cos = 0 3θ = 2nπ ± θ
2 2
θ − α θ−α Taking (+) sign Taking (–) sign
sin = 0 ⇒ = nπ ⇒ θ = 2nπ + α
2
2 3θ = 2nπ + θ 3θ = 2nπ – θ
θ = –4π + α, 2π + α, 4π + α, α, –2π + α, .... 2θ = 2nπ 4θ = 2nπ
, nπ 1 3
θ = nπ θ = − 3a ×
+ 2b = 0
2 2 2
3π 3a b 1
π
, π, 3b = ⇒ = fnf.
0, π, 2π, ... 0, , ... 2 a 2
2 2
Sushmita:
π 3π
–π, –2π, ... − , –π, − , ... 2 x x
2 2 1 − tan 2 tan
3a 2 + 2b 2
= c
2 x 2 x
nπ 1 + tan 1 + tan
Union: θ = n Є I fnf. 2 2
2
x
Question
eku yks% tan = t
2
Q. Solve following equation: sin 9θ = sinθ
3a − 3at 2 + 4bt = c + ct2
Sol. 9θ = nπ + (–1)nθ
If n ≡ even α
tan
9θ = nπ + θ ⇒ 8θ = nπ 2
nπ
(c + )
3a t 2 − 4bt + c − 3a =0
θ = , n є even; put n = 2m, m ∈ I β
8 tan
2
2mπ mπ
θ = = α β 1
8 4 tan + =
.
If n ≡ odd 2 2 3
9θ = nπ – θ ⇒ 10θ = nπ −4b
α β −
nπ tan
+ tan c + 3a
, n ∈ odd; put n = 2m + 1, m ∈ I 2 2 1
θ = ⇒ = . =
10 α β 3 c − 3a
1 − tan tan
π 2 2 1 −
θ = (2m + 1) c + 3a
10
mπ π b 1
fnf ⇒ θ є ∪ (2m + 1) , m Є I. ⇒ = fnf
2 2
4 10
Question [JEE (Adv.)-2018] Question
Q. Let a, b, c be three non zero numbers such Q. Solve following equation:
( ) + ( tan θ + 3 )
3a cos x + 2b sin x = c, 2 2
that the equation
2 sin θ − 3 = 0
π π
x є − , has two distinct real roots 3
2 2 Sol. Sunita: sin θ = & tan θ = − 3
2
π
α and β with α + β = . Then the value of π
3 sin θ = sin
3
b
is π
a θ = nπ + ( −1)
n
...(1)
3
Sol. Sunita: 3a cos α + 2b sin α =c ...(1)
2 π 7 π −4 π −5 π
θ= , , , , ... and tan θ = − 3
3a cos β + 2b sin β =c ...(2) 3 3 3 3
Subtract π π
tan θ = tan − ⇒ θ = mπ − ...(2)
3a {cos α − cos β} + 2b{sin α − sin β} = 0 3 3
( α − β) ( α + β) α +β α − β π 2π 4π 5 π 8π
3a −2 sin sin + 2b 2 cos
sin
= 0 θ= − , ,− , , .
2 2
2 2 3 3 3 3 3
2
Mathematics
, ⇒ from (1) ∩ (2) Question
4 π 2 π 8π 14 π
θ = − , , , , ... Q. Solve: tan x tan 4x = 1
3 3 3 3
1
π Sol. tan 4x = 1
θ = (6n – 4) , n ∈ I fnf. cot x
3
π
3 tan 4x = cot x ⇒ tan 4x = tan − x
Sushmita: sin θ = ∩ tan θ = − 3 2
2
2π π 2nπ + π
2π 120° 4x = nπ + − x ⇒ 5x =
⇒θ= 3 2 2
3
π
⇒ Now to Generalise add (2np): x = (2n + 1) , n ≡ Int.
10
2π π π 3π 5 π 7 π 9π 11π 13π 15 π
θ 2nπ +
= = (6n + 2) , n ∈ I fnf. x = , , , , , , , , ...
3 3 10 10 10 10 10 10 10 10
−π −3π
, , ...
DHYAAN RKHNA 10 10
π π
x ≠ ( 2n + 1) & x ≠ ( 2n + 1)
2 8
# For whole Mathematics
fnf ⇒ x Є (2n + 1) π − (2n + 1) π , n Є I
1. Avoid squaring: In case of squaring final 10 2
solution be checked.
4. Aapka answer, dikhne mein alag ho skta hai.
Squaring se extra solutions aa skte hai.
2. Avoid cancellation in multiplication. 1
Ex: sin x = −
Cancellation leads to loss of original solutions. 2
If ab = bc ⇒ a = c π 7π
sin x = sin − or sin x = sin .
⇒ b(a – c) = 0 ⇒(a = c) ∪ (b = 0) 6 6
b = 0 (or) U a – c = 0 π n 7π
x = nπ + (–1)n + 1 or x = nπ + ( −1)
b c 6 6
If = ⇒ b = c
a a both are Same
If a + b = a + c ⇒ b = c
3. Answer shouldn’t contain values for which
Note
any of given function is not defined.
Sin, cos, tan → +ve value then a ∈ I quad.
(tan; cot; cosec; sec) vxj esa present gks rks lroZQ
jgs lko/ku jgsaA Sin → –ve value then take a ∈ IV quad.
Ex.: Solve: tan 3x – tan 2x = 1 + tan 3x tan2x cos → –ve value then take a ∈ II quad.
tan → –ve value then take a ∈ IV quad.
tan 3x − tan 2x
Sol. = 1
1 + tan 3x tan 2x
THOUGHT PROCESS
π π
tan (3x – 2x) = 1 = tan ⇒ x = nπ + # T
ry to make fucntion and argument (angle)
4 4 same.
π 5 π 9π # F
actorize into linear factors and write
x= , , , ...
4 4 4 general solution.
tan2x, not defined for all
3π −7 π # A
gr factorization na ho to perfect square
− , , ... banane ka socho.
4 4
so x є f (No solution) # Look for an identity.
3
Trigonometric Equations
GENERAL SOLUTIONS OF T.E. θ + α
Or cos = 0 ⇒
θ+ α
= (2n + 1)
π
2 2 2
Types of Solutions: θ = (2n + 1)π – α
General solution: All angles π – α, 3π – α, 5π – α, – π – α ...
Principle solution: θ Є [0, 2π)
Particular solution: All solutions in the given θ = nπ + (–1)nα, n Є I
interval.
Note Proof:
sin x = 0 ↔ x = nπ cos θ = cos α ⇒ cos θ – cos α = 0
π θ + α θ − α
cos x = 0 ↔ x = (2n + 1)
2 −2 sin sin = 0
2 2
π
sin x = 1 ↔ x = (4n + 1) θ = 2nπ – α (or) θ = 2nπ + α
2
π ⇒ θ = 2nπ ± α
sin x = –1 ↔ x = (4n – 1)
2
cos x = 1 ↔ x = 2nπ Question
cos x = –1 ↔ x = (2n – 1)π
(Where 'n' is any integer) Q. Solve following equation: tan 3θ = –1
π
Sol. Sumit: tan 3θ = tan −
ONLY FOUR FORMULA'S 4
π π(4n − 1)
3θ = nπ − =
4 4
(1) sin θ = sin α ⇒ θ = nπ + (–1)nα, (n Є I)
π
(2) cos θ = cos α ⇒ θ = 2nπ ± α, (n Є I) θ = (4n − 1)
12 Same
(3) tan θ = tan α ⇒ θ = nπ + α, (n Є I) 3π
(4) sin2 θ = sin2 α Santosh: tan 3θ = tan
4
cos2 θ = cos2 α ⇒ θ = nπ ± α, (n Є I) 3π π
3θ = nπ + ⇒ θ = (4n + 3)
tan2 θ = tan2 α 4 12
Question
Proof:
Q. Solve following equation: cos 3θ = cos θ
sin θ = sin α ⇒ sin θ – sin α = 0
Sol. cos 3θ = cos θ
θ − α θ + α
2 sin cos = 0 3θ = 2nπ ± θ
2 2
θ − α θ−α Taking (+) sign Taking (–) sign
sin = 0 ⇒ = nπ ⇒ θ = 2nπ + α
2
2 3θ = 2nπ + θ 3θ = 2nπ – θ
θ = –4π + α, 2π + α, 4π + α, α, –2π + α, .... 2θ = 2nπ 4θ = 2nπ
, nπ 1 3
θ = nπ θ = − 3a ×
+ 2b = 0
2 2 2
3π 3a b 1
π
, π, 3b = ⇒ = fnf.
0, π, 2π, ... 0, , ... 2 a 2
2 2
Sushmita:
π 3π
–π, –2π, ... − , –π, − , ... 2 x x
2 2 1 − tan 2 tan
3a 2 + 2b 2
= c
2 x 2 x
nπ 1 + tan 1 + tan
Union: θ = n Є I fnf. 2 2
2
x
Question
eku yks% tan = t
2
Q. Solve following equation: sin 9θ = sinθ
3a − 3at 2 + 4bt = c + ct2
Sol. 9θ = nπ + (–1)nθ
If n ≡ even α
tan
9θ = nπ + θ ⇒ 8θ = nπ 2
nπ
(c + )
3a t 2 − 4bt + c − 3a =0
θ = , n є even; put n = 2m, m ∈ I β
8 tan
2
2mπ mπ
θ = = α β 1
8 4 tan + =
.
If n ≡ odd 2 2 3
9θ = nπ – θ ⇒ 10θ = nπ −4b
α β −
nπ tan
+ tan c + 3a
, n ∈ odd; put n = 2m + 1, m ∈ I 2 2 1
θ = ⇒ = . =
10 α β 3 c − 3a
1 − tan tan
π 2 2 1 −
θ = (2m + 1) c + 3a
10
mπ π b 1
fnf ⇒ θ є ∪ (2m + 1) , m Є I. ⇒ = fnf
2 2
4 10
Question [JEE (Adv.)-2018] Question
Q. Let a, b, c be three non zero numbers such Q. Solve following equation:
( ) + ( tan θ + 3 )
3a cos x + 2b sin x = c, 2 2
that the equation
2 sin θ − 3 = 0
π π
x є − , has two distinct real roots 3
2 2 Sol. Sunita: sin θ = & tan θ = − 3
2
π
α and β with α + β = . Then the value of π
3 sin θ = sin
3
b
is π
a θ = nπ + ( −1)
n
...(1)
3
Sol. Sunita: 3a cos α + 2b sin α =c ...(1)
2 π 7 π −4 π −5 π
θ= , , , , ... and tan θ = − 3
3a cos β + 2b sin β =c ...(2) 3 3 3 3
Subtract π π
tan θ = tan − ⇒ θ = mπ − ...(2)
3a {cos α − cos β} + 2b{sin α − sin β} = 0 3 3
( α − β) ( α + β) α +β α − β π 2π 4π 5 π 8π
3a −2 sin sin + 2b 2 cos
sin
= 0 θ= − , ,− , , .
2 2
2 2 3 3 3 3 3
2
Mathematics
, ⇒ from (1) ∩ (2) Question
4 π 2 π 8π 14 π
θ = − , , , , ... Q. Solve: tan x tan 4x = 1
3 3 3 3
1
π Sol. tan 4x = 1
θ = (6n – 4) , n ∈ I fnf. cot x
3
π
3 tan 4x = cot x ⇒ tan 4x = tan − x
Sushmita: sin θ = ∩ tan θ = − 3 2
2
2π π 2nπ + π
2π 120° 4x = nπ + − x ⇒ 5x =
⇒θ= 3 2 2
3
π
⇒ Now to Generalise add (2np): x = (2n + 1) , n ≡ Int.
10
2π π π 3π 5 π 7 π 9π 11π 13π 15 π
θ 2nπ +
= = (6n + 2) , n ∈ I fnf. x = , , , , , , , , ...
3 3 10 10 10 10 10 10 10 10
−π −3π
, , ...
DHYAAN RKHNA 10 10
π π
x ≠ ( 2n + 1) & x ≠ ( 2n + 1)
2 8
# For whole Mathematics
fnf ⇒ x Є (2n + 1) π − (2n + 1) π , n Є I
1. Avoid squaring: In case of squaring final 10 2
solution be checked.
4. Aapka answer, dikhne mein alag ho skta hai.
Squaring se extra solutions aa skte hai.
2. Avoid cancellation in multiplication. 1
Ex: sin x = −
Cancellation leads to loss of original solutions. 2
If ab = bc ⇒ a = c π 7π
sin x = sin − or sin x = sin .
⇒ b(a – c) = 0 ⇒(a = c) ∪ (b = 0) 6 6
b = 0 (or) U a – c = 0 π n 7π
x = nπ + (–1)n + 1 or x = nπ + ( −1)
b c 6 6
If = ⇒ b = c
a a both are Same
If a + b = a + c ⇒ b = c
3. Answer shouldn’t contain values for which
Note
any of given function is not defined.
Sin, cos, tan → +ve value then a ∈ I quad.
(tan; cot; cosec; sec) vxj esa present gks rks lroZQ
jgs lko/ku jgsaA Sin → –ve value then take a ∈ IV quad.
Ex.: Solve: tan 3x – tan 2x = 1 + tan 3x tan2x cos → –ve value then take a ∈ II quad.
tan → –ve value then take a ∈ IV quad.
tan 3x − tan 2x
Sol. = 1
1 + tan 3x tan 2x
THOUGHT PROCESS
π π
tan (3x – 2x) = 1 = tan ⇒ x = nπ + # T
ry to make fucntion and argument (angle)
4 4 same.
π 5 π 9π # F
actorize into linear factors and write
x= , , , ...
4 4 4 general solution.
tan2x, not defined for all
3π −7 π # A
gr factorization na ho to perfect square
− , , ... banane ka socho.
4 4
so x є f (No solution) # Look for an identity.
3
Trigonometric Equations
