(MATHEMATICS) TRIGONOMETRIC EQUATION

1. TRIGONOMETRIC EQUATION:
An equation involving one or more trigonometrical ratios of unknown angles is called a
trigonometrical equation.
2. SOLUTION OF TRIGONOMETRIC EQUATION:
A value of the unknown angle which satisfies the given equation is called a solution of the
trigonometric equation.
(a) Principal solution: The solution of the trigonometric equation lying in the interval
(0, 2𝜋)
(b) General solution: Since all the trigonometric functions are many one & periodic, hence
there are infinite values of 𝜃 for which trigonometric functions have the same value.
All such possible values of 𝜃 for which the given trigonometric function is satisfied is given
by a general formula. Such a general formula is called general solution of trigonometric
equation.
(c) Particular solution: The solution of the trigonometric equation lying in the given
interval.
3. GENERAL SOLUTIONS OF SOME TRIGONOMETRIC EQUATIONS (TO BE REMEMBERED):
(a) If sin θ = 0, then θ = nπ, n ∈ I (set of integers)
π
(b) If cosθ = 0, then θ = (2n + 1) 2 , n ∈ I
(c) If tanθ = 0, then θ = nπ, n ∈ I
−π π
(d) If sinθ = sin α, then θ = nπ + (−1)n α, where α ∈ [ 2 , 2 ] , n ∈ I
(e) If cosθ = cos α, then θ = 2 nπ ± α, n ∈ I, α ∈ [0, π]
−π π
(f) If tanθ = tan α, then θ = nπ + α, n ∈ I, α ∈ ( 2 , 2 )
π π
(g) If sinθ = 1, then θ = 2nπ + 2 = (4n + 1) 2 , n ∈ I
(h) If cosθ = 1, then θ = 2nπ, n ∈ I
(i) If sin2 θ = sin2 α or cos2 θ = cos2 α or tan2 θ = tan2 α, then θ = nπ ± α, n ∈ I
(j) For n ∈ I, sin nπ = 0 and cos nπ = (−1)n , n ∈ I
sin (nπ + θ) = (−1)n sin θ cos (nπ + θ) = (−1)n cos θ
n−1
nπ nπ
(k) cos nπ = (−1)n , n ∈ I If n is an odd integer, then sin = (−1) 2 , cos = 0,
2 2
n−1
nπ
sin ( 2 + θ) = (−1) 2 cosθ
n+1
nπ
cos ( + θ) = (−1) 2 sinθ
2
tan 3x − tan2x
Illustration 1: Find the set of values of x for which 1+tan3x⋅tan2x = 1
tan 3x − tan 2x
Solution : We have, 1 + tan 3x⋅tan 2x = 1 ⇒ tan(3x − 2x) = 1 ⇒ tan x = 1
π π
⇒ tan x = tan 4 ⇒ x = nπ + 4 , n ∈ I { using tan θ = tan α ⇔ θ = nπ + α}
But for this value of x, tan 2x is not defined. Hence the solution set for x is ϕ.Ans.

Do yourself-1:
(i) Find general solutions of the following equations:
1 3θ 3θ
(a) sin θ = 2 (b) cos ( 2 ) = 0 (c) tan ( 4 ) = 0
θ
(d) cos2 2θ = 1 (e) √3sec 2θ = 2 (f) cosec (2) = −1

APNI KAKSHA 1
(MATHEMATICS) TRIGONOMETRIC EQUATION
4. IMPORTANT POINTS TO BE REMEMBERED WHILE SOLVING TRIGONOMETRIC
EQUATIONS:
(a) For equations of the type sin θ = k or cos θ = k, one must check that |k| ≤ 1.
(b) Avoid squaring the equations, if possible, because it may lead to extraneous solutions.
Reject extra solutions if they do not satisfy the given equation.
(c) Do not cancel the common variable factor from the two sides of the equations which are
in a product because we may loose some solutions.
(d) The answer should not contain such values of θ, which make any of the terms undefined
or infinite.
(i) Check that denominator is not zero at any stage while solving equations.
π
(ii) If tan θ or sec θ is involved in the equations, θ should not be odd multiple of 2 .

(iii) If cot θ or cosec θ is involved in the equation, θ should not be multiple of π or 0.

5. DIFFERENT STRATEGIES FOR SOLVING TRIGONOMETRIC EQUATIONS:
(a) Solving trigonometric equations by factorization.
e.g. (2sin x − cos x)(1 + cos x) = sin2 x
∴ (2sin x − cos x)(1 + cos x) − (1 − cos 2 x) = 0
∴ (1 + cos x)(2sin x − cos x − 1 + cos x) = 0 ∴ (1 + cos x)(2sin x − 1) = 0
1
⇒ cos x = −1 or sin x = 2 ⇒ cos x = −1 = cos π ⇒ x = 2nπ + π = (2n + 1)π, n ∈ I
1 π π
or sin x = 2 = sin ⇒ x = kπ + (−1)k 6 , k ∈ I Ans.
6
1
Illustration 2: If 6 sin θ, cos θ and tan θ are in G.P. then the general solution for θ is-
π π π
(A) 2nπ ± 3 (B) 2nπ ± 6 (C) nπ ± 3 (D) none of these
1 1
Solution: Since, 6 sin θ, cos θ, tan θ are in G.P. ⇒ cos 2 θ = 6 sinθ ⋅ tanθ ⇒ 6cos3 θ + cos 2 θ − 1 = 0

∴ (2cos θ − 1)(3cos2 θ + 2cos θ + 1) = 0

1
⇒ cos θ = 2 (other values of cos θ are imaginary)
π π
⇒ cos θ = cos ⇒ θ = 2nπ ± 3 , n ∈ I. Ans.
3

(b) Solving of trigonometric equation by reducing it to a quadratic equation.

e.g. 6 − 10cosx = 3sin2 x
∴ 6 − 10cosx = 3 − 3cos 2 x ⇒ 3cos 2 x − 10cos x + 3 = 0
1
⇒ (3cos x − 1)(cos x − 3) = 0 ⇒ cos x = 3 or cos x = 3

Since cos x = 3 is not possible as − 1 ≤ cos x ≤ 1

1 1 1
∴ cos x = 3 = cos (cos −1 3) ⇒ x = 2nπ ± cos−1 (3) , n ∈ I Ans.

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(MATHEMATICS) TRIGONOMETRIC EQUATION
1
Illustration 3: Solve sin2 θ − cos θ = 4, for θ and write the values of θ in the interval 0 ≤ θ ≤ 2π.
1
Solution: The given equation can be written as 1 − cos2 θ − cosθ = 4 ⇒ cos 2 θ + cosθ − 3/4 = 0
1 3
⇒ 4cos2 θ + 4cosθ − 3 = 0 ⇒ (2cosθ − 1)(2cosθ + 3) = 0 ⇒ cosθ = 2 , − 2

Since, cosθ = −3/2 is not possible as −1 ≤ cosθ ≤ 1

1 π π
∴ cosθ = 2 ⇒ cosθ = cos ⇒ θ = 2nπ ± 3 , n ∈ I
3
π 5π
For the given interval, n = 0 and n = 1. ⇒ θ = 3 , Ans.
3

Illustration 4: Find the number of solutions of tan x + sec x = 2 cos x in [0,2π].

Solution: Here, tan x + sec x = 2cos x ⇒ sin x + 1 = 2 cos2 x ⇒ 2 sin2 x + sin x − 1 = 0
1 3π
⇒ sin x = 2 , −1 But sin x = −1 ⇒ x = for which tan x + sec x = 2cos x is not defined.
2
1 π 5π
Thus sin x = 2 ⇒ x = 6 , ⇒ number of solutions of tan x + sec x = 2cos x is 2. Ans.
6

Illustration 5: Solve the equation 5sin2 x − 7sin xcos x + 16cos2 x = 4

Solution: To solve this equation we use the fundamental formula of trigonometric identities,
sin2 x + cos2 x = 1 writing the equation in the form.
5 sin2 x − 7sin x ⋅ cos x + 16cos 2 x = 4(sin2 x + cos 2 x)
⇒ sin2 x − 7sinx cos x + 12 cos2 x = 0
dividing by cos 2 x on both side we get, tan2 x − 7tanx + 12 = 0
Now it can be factorized as : (tan x − 3)(tan x − 4) = 0 ⇒ tan x = 3,4
i.e., tanx = tan (tan−1 3) or tan x = tan (tan−1 4)
x = nπ + tan−1 3 or x = nπ + tan−1 4, n ∈ I.Ans.
n 2 x−3 sin x+2
Illustration 6: If x ≠ 2 , n ∈ I and (cos x)sin = 1, then find the general solutions of x.
nπ 2 x−3 sin x+2
Solution : As x ≠ ⇒ cos x ≠ 0,1, −1 So, (cos x)sin =1
2
⇒ sin2 x − 3sin x + 2 = 0
sin x = 1, 2 where sin x = 2 is not possible and
∴ (sin x − 2)(sin x − 1) = 0 ⇒ ̅̅̅̅̅̅
nπ
sin x = 1 which is also not possible as x ≠ 2 ∴ no general solution is possible. Ans.
7
Illustration 7: Solve the equation sin4 x + cos4 x = 2 sin x ⋅ cos x
7 7
Solution : sin4 x + cos 4 x = 2 sin x ⋅ cos x ⇒ (sin2 x + cos2 x)2 − 2 sin2 x cos2 x = 2 sin x ⋅ cosx
1 7
⇒ 1 − (sin 2x)2 = (sin 2x) ⇒ 2sin2 2x + 7sin 2x − 4 = 0
2 4
1
⇒ (2 sin 2x − 1)(sin 2x + 4) = 0 ⇒ sin 2x = 2 or sin 2x = −4
π
(which is not possible) ⇒ 2x = nπ + (−1)n , n ∈ I
6
nπ π
i.e. x = + (−1)n 12 , n ∈ I Ans.
2

APNI KAKSHA 3
(MATHEMATICS) TRIGONOMETRIC EQUATION
Do yourself-2:
(i) Solve the following equations:

(a) 3sin x + 2cos2 x = 0 (b) sec 2 2α = 1 − tan 2α

(c) 7cos2 θ + 3sin2 θ = 4 (d) 4cos θ − 3sec θ = tan θ

(ii) Solve the equation: 2sin2 θ + sin2 2θ = 2 for θ ∈ (−π, π).

(c) Solving trigonometric equations by introducing an auxiliary argument.

Consider, asin θ + bcos θ = c ………….(i)

a b c
∴ sin θ + cos θ =
√a2 + b 2 √a2 + b 2 √a2 + b 2

equation (i) has a solution only if |c| ≤ √a2 + b 2

a b b
let √a2 = cos ϕ, √a2 = sin ϕ&ϕ = tan−1
+b2 +b2 a
by introducing this auxiliary argument ϕ, equation (i) reduces to
c
sin (θ + ϕ) = √a2 Now this equation can be solved easily.
+b2

Illustration 8: Find the number of distinct solutions of

sec x + tan x = √3, where 0 ≤ x ≤ 3π.
Solution: Here, sec x + tan x = √3

⇒ 1 + sin x = √3cos x or √3cos x − sin x = 1

dividing both sides by √a2 + b 2 i.e. √4 = 2 we get

√3 1 1 π π 1
⇒ cos x − 2 sin x = 2 ⇒ cos cos x − sin sin x = 2 ⇒
2 6 6
π 1
cos (x + 6 ) = 2

As 0 ≤ x ≤ 3π

π π π π π 5π 7π π 3π 13π
≤ x + ≤ 3π + ⇒ x + = , , ⇒x= , ,
6 6 6 6 3 3 3 6 2 6
3π
But at x = , tan x and sec x is not defined ∴ Total number of solutions are 2.Ans.
2

Illustration 9: Prove that the equation kcos x − 3sin x = k + 1 possess a solution iff k ∈ (−∞, 4].
k 3 k+1
Solution: Here, k cosx − 3sinx = k + 1, could be re-written as: √k2 cos x − √k2 sin x = √k2
+9 +9 +9
k+1 3 k+1
or cos (x + ϕ) = √k2 , where tanϕ = k which possess a solution only if −1 ≤ √k2 ≤1
+9 +9
k+1
i.e., |√k2 | ≤ 1 i.e., (k + 1)2 ≤ k 2 + 9 i.e., k 2 + 2k + 1 ≤ k 2 + 9 or k ≤ 4
+9

⇒ The interval of k for which the equation (kcos x − 3sin x = k + 1)

has a solution is (−∞, 4]. Ans.

APNI KAKSHA 4
(MATHEMATICS) TRIGONOMETRIC EQUATION
Do yourself - 3:
(i) Solve the following equations:
(a) sin x + √2 = cos x. (b) cosec θ = 1 + cot θ
(d) Solving trigonometric equations by transforming sum of trigonometric functions into
product.
e.g. cos 3x + sin 2x − sin 4x = 0
1
cos 3x − 2sin xcos 3x = 0 ⇒ (cos 3x)(1 − 2sin x) = 0 ⇒ cos 3x = 0 or sin x = 2
π 1 π
⇒ cos 3x = 0 = cos sin x = = sin
2 2 6
π π 2nπ π
⇒ 3x = 2nπ ± 2 x = mπ + (−1)m 6 ; (n, m ∈ I) ⇒ x = ± 6 Ans.
3

Illustration 10: Solve: cos θ + cos 3θ + cos 5θ + cos 7θ = 0

Solution: We have cos θ + cos 7θ + cos 3θ + cos 5θ = 0
⇒ 2cos 4θcos 3θ + 2cos 4θcos θ = 0
⇒ cos 4θ(cos 3θ + cos θ) = 0 ⇒ cos 4θ(2cos 2θcos θ) = 0
⇒Either cos θ = 0 ⇒ θ = (2n1 + 1)π/2, n1 ∈ I
π
or cos 2θ = 0 ⇒ θ = (2n2 + 1) 4 , n2 ∈ I
π
or cos 4θ = 0 ⇒ θ = (2n3 + 1) 8 , n3 ∈ I Ans.

(e) Solving trigonometric equations by transforming a product into sum.

e.g. sin 5x ⋅ cos 3x = sin 6x ⋅ cos 2x
sin 8x + sin 2x = sin 8x + sin 4x ∴ 2sin 2x ⋅ cos 2x − sin 2x = 0
1
⇒ sin 2x(2cos 2x − 1) = 0 ⇒ sin 2x = 0 or cos 2x = 2
1 π
⇒ sin 2x = 0 = sin 0 cos 2x = 2 = cos 3
π
⇒ 2x = nπ + (−1)n × 0, n ∈ I 2x = 2 mπ ± , m ∈ I
3
nπ π
⇒ x= ,n ∈ I x = mπ ± 6 , m ∈ I
2
1
Illustration 11: Solve: cos θcos 2θcos 3θ = 4; where 0 ≤ θ ≤ π.
1 1 1
Solution : (2cos θcos 3θ)cos 2θ = 4 ⇒ (cos 2θ + cos 4θ)cos 2θ = 2
2
1 1
⇒ 2 [2cos 2 2θ + 2cos 4θcos 2θ] = 2 ⇒ 1 + cos 4θ + 2cos 4θcos 2θ = 1

∴ cos 4θ(1 + 2cos 2θ) = 0

cos 4θ = 0 or (1 + 2cos 2θ) = 0
Now from the first equation: 2cos 4θ = 0 = cos (π/2)
1 π
∴ 4θ = (n + ) π ⇒ θ = (2n + 1) , n ∈ I
2 8

APNI KAKSHA 5
(MATHEMATICS) TRIGONOMETRIC EQUATION
π 3π 5π 7π
for n = 0, θ = 8 ; n = 1, θ = ; n = 2, θ = ; n = 3, θ = (∴ 0 ≤ θ ≤ π) and
8 8 8
from the second equation:
1
cos 2θ = − = −cos (π/3) = cos (π − π/3) = cos (2π/3)
2
2π π
∴ 2θ = 2kπ ± ⇒ θ = kπ ± , k ∈ I
3 3
π 2π π π 3π 5π 2π 7π
again for k = 0, θ = 3 ; k = 1, θ = 3 (∵ 0 ≤ θ ≤ π) ∴ θ = 8 , 3 , 8 , 8 , 3 , 8 Ans.
Do yourself-4:
(i) Solve 4sin θsin 2θsin 4θ = sin 3θ (ii) Solve for x: sin x + sin 3x + sin 5x = 0
(f) Solving equations by a change of variable:
(i) Equations of the form P(sin x ± cos x, sin x ⋅ cos x) = 0, where P(y, z) is a polynomial,
can be solved by the substitution: cos x ± sin x = t ⇒ 1 ± 2sin x ⋅ cos x = t 2
Illustration 12: Solve: sin x + cos x = 1 + sin x ⋅ cos x.
Solution : put sin x + cos x = t ⇒ sin2 x + cos2 x + 2 sin x ⋅ cos x = t 2
t2 − 1
⇒ 2sin xcos x = t 2 − 1 (∵ sin2 x + cos 2 x = 1) ⇒ sin xcos x = ( )
2
Substituting above result in given equation, we get:
t2 − 1
t=1+ ⇒ 2t = t 2 + 1 ⇒ t 2 − 2t + 1 = 0
2
⇒ (t − 1)2 = 0 ⇒ t = 1 ⇒ sin x + cos x = 1
Dividing both sides by √12 + 12 i.e. √2. we get
1 1 π 1 π 1
⇒ sin x + cos x = ⇒ cos xcos
+ sin x ⋅ sin =
√2 √2 √2 4 4 √2
π π π π
⇒ cos (x − ) = cos ⇒ x − = 2nπ ±
4 4 4 4
π π
⇒ x = 2nπ or x = 2nπ + 2 = (4n + 1) 2 , n ∈ I Ans.

(ii) Equations of the form of a sin x + bcos x + d = 0, where a, b & d are real numbers can
be solved by changing sin x&cos x into their corresponding tangent of half the angle.
Illustration 13 : Solve : 3cos x + 4sin x = 5
x x
1−tan2 x/2 2tan x/2 3−3tan2 8tan
Solution : ⇒ 3 (1+tan2 x/2) + 4 (1+tan2 x/2) = 5 ⇒ x
2
+ 2
x =5
1+tan2 1+tan2
2 2
x x x x x
⇒ 3 − 3tan2 + 8tan = 5 + 5tan2 ⇒ 8tan2 − 8tan +2 =0
2 2 2 2 2

x x x 2
⇒ 4tan2 − 4tan + 1 = 0 ⇒ (2tan − 1) = 0
2 2 2
x x 1 1
⇒ 2tan (2) − 1 = 0 ⇒ tan = 2 = tan (tan−1 2)
2
x 1 1
⇒ 2 = nπ + tan−1 (2) , n ∈ I ⇒ x = 2nπ + 2 tan−1 (2) , n ∈ IAns.

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(MATHEMATICS) TRIGONOMETRIC EQUATION
(g) Solving trigonometric equations with the use of the boundness of the functions
involved.

Illustration 14: Solve the equation (sin x + cos x)1+sin 2x = 2, when 0 ≤ x ≤ π

Solution : We know, −√a2 + b 2 ≤ asin θ + bcos θ ≤ √a2 + b 2 and −1 ≤ sin θ ≤ 1.
∴ (sin x + cos x) admits the maximum value as √2 and (1 + sin 2x) admits the
maximum value as 2. Also (√2)2 = 2.
∴ the equation could hold only when, sin x + cos x = √2 and 1 + sin 2x = 2
Now, sin x + cos x = √2
π
⇒ cos (x − 4 ) = 1 ⇒ x = 2nπ + π/4, n ∈ I ………..(i)
π
and 1 + sin 2x = 2 ⇒ sin 2x = 1 = sin 2
mπ mπ π
⇒ 2x = mπ + (−1) ,m∈ I⇒x= + (−1)m 4 ………..(ii)
2 2
π
The value of x in [0, π] satisfying equations (i) and (ii) is x = 4

(when n = 0& m = 0) Ans.

Note: sin x + cos x = −√2 and 1 + sin 2x = 2 also satisfies but as x ≥ 0, this solution is not in
domain.
1
Illustration 15: Solve for x and y: 2cos2 x √y 2 − y + 1/2 ≤ 1
1
Solution: 2cos2 x √y 2 − y + 1/2 ≤ 1 …………(i)
1
1 2 1 2
2cos2 x √(y − 2) + (2) ≤ 1

1
1 2 1 2 1
Minimum value of 2cos2 x = 2 ⇒ Minimum value of √(y − 2) + (2) = 2
1
1
⇒ Minimum value of 2cos2 x √y 2 − y + is 1⇒ (i) is possible
2

1
1 2 1 2
when2cos2 x √(y − 2) + (2) ≤ 1 ⇒ cos 2 x = 1 and y = 1/2

⇒ cos x = ±1 ⇒ x = nπ, where n ∈ I Hence x = nπ, n ∈ I and y = 1/2. Ans.

x 1 π
Illustration 16: The number of solution(s) of 2cos2 (2) sin2 x = x 2 + x2 , 0 ≤ x ≤ 2 is/are-
(A) 0 (B) 1 (C) infinite (D) none of these
x 1 1
Solution : Let y = 2cos2 (2) sin2 x = x 2 + x2 ⇒ y = (1 + cos x)sin2 x and y = x 2 + x2
when y = (1 + cos x)sin2 x = ( a number < 2)( a number ≤ 1) ⇒ y < 2
1 1 2
and when y = x 2 + x2 = (x − x) + 2 ≥ 2 ⇒ y ≥ 2
No value of y can be obtained satisfying (i) and (ii), simultaneously
⇒ No real solution of the equation exists. Ans.
APNI KAKSHA 7
(MATHEMATICS) TRIGONOMETRIC EQUATION
Note: If L.H.S. of the given trigonometric equation is always less than or equal to k and RHS is always
greater than k, then no solution exists. If both the sides are equal to k for same value of θ, then
solution exists and if they are equal for different values of θ, then solution does not exist.
Do yourself-5 :
(i) If x 2 − 4x + 5 − sin y = 0, y ∈ [0,2π), then –
(A) x = 1, y = 0 (B) x = 1, y = π/2
(C) x = 2, y = 0 (D) x = 2, y = π/2
1
(ii) If sin x + cos x = √y + y , y > 0, x ∈ [0, π], then find the least positive value of x satisfying the
given condition.
6. TRIGONOMETRIC INEQUALITIES:
There is no general rule to solve trigonometric in equations and the same rules of algebra are
valid here. The domain and range of trigonometric functions should be kept in mind.
Illustration 17: Find the solution set of inequality sin x > 1/2.
1
Solution: When sin x = 2, the two values of x between 0 and 2π are π/6 and 5π/6.
1
From the graph of y = sin x, it is obvious that between 0 and 2π, sin x > 2
for π/6 < x < 5π/6 Hence, sin x > 1/2 ⇒ 2nπ + π/6 < x < 2nπ + 5π/6, n ∈ I

π 5π
Thus, the required solution set is ∪ (2nπ + 6 , 2nπ + ) Ans.
n∈I 6

Illustration 18: Find the values of α lying between 0 and π for which the inequality:

tan α > tan3 α is valid.

Solution : We have : tan α − tan3 α > 0 ⇒ tan α(1 − tan2 α) > 0

⇒ (tan α)(tan α + 1)(tan α − 1) < 0 So tan α < −1,0 < tan α < 1
π π 3π
∴ Given inequality holds for α ∈ (0, 4 ) ∪ ( 2 , ) Ans.
4

Do yourself - 6:

(i) Find the solution set of the inequality: cos x ≥ −1/2.

(ii) Find the values of x in the interval [0,2π] for which 4sin2 x − 8sin x + 3 ≤ 0.

Miscellaneous Illustration:

Illustration 19: Solve the following equation: tan2 θ + sec 2 θ + 3 = 2(√2sec θ + tan θ)

Solution : We have tan2 θ + sec 2 θ + 3 = 2√2sec θ + 2tan θ

⇒ tan2 θ − 2tan θ + sec 2 θ − 2√2sec θ + 3 = 0
⇒ tan2 θ + 1 − 2tan θ + sec 2 θ − 2√2sec θ + 2 = 0
APNI KAKSHA 8
(MATHEMATICS) TRIGONOMETRIC EQUATION
⇒ (tan θ − 1)2 + (sec θ − √2)2 = 0 ⇒ tan θ = 1 and sec θ = √2
As the periodicity of tan θ and sec θ are not same, we get
π
θ = 2nπ + 4 , n ∈ I Ans.

Illustration 20 : Find the solution set of equation 5(1+log5 cos x) = 5/2.

Solution : Taking log to base 5 on both sides in given equation :

(1 + log 5 cos x) ⋅ log 5 5 = log 5 (5/2) ⇒ log 5 5 + log 5 cos x = log 5 5 − log 5 2

⇒ log 5 cos x = −log 5 2 ⇒ cos x = 1/2 ⇒ x = 2nπ ± π/3, n ∈ I Ans.

π π
Illustration 21: If the set of all values of x in (− 2 , 2 )
aπ bπ
satisfying |4sin x + √2| < √6 is (24 , 24 )

a−b
then find the value of | |.
3

Solution : |4sin x + √2| < √6

⇒ −√6 < 4sin x + √2 < √6

⇒ −√6 − √2 < 4sin x < √6 − √2

−(√6+√2) √6−√2
⇒ < sin x <
4 4

5π π π π
⇒ − 12 < x < 12 for x ∈ (− 2 , 2 )

aπ bπ
Comparing with 24 < x < , we get, a = −10, b = 2
24

a−b −10−2
∴| |=| |=4 Ans.
3 3

Illustration 22 : The number of values of x in the interval [0,5𝜋] satisfying the equation.

3sin2 𝑥 − 7sin 𝑥 + 2 = 0 is- [JEE 98]

(A) 0 (B) 5 (C) 6 (D) 10

Solution : 3sin2 x − 7sin x + 2 = 0

⇒ (3sin x − 1)(sin x − 2) = 0

∵ sin x ≠ 2
1
⇒ sin x = 3 = sin α( say)

where α is the least positive value of

APNI KAKSHA 9
(MATHEMATICS) TRIGONOMETRIC EQUATION
1
x such that sin α = 3.
π
Clearly 0 < α < 2 . We get the solution,

x = α, π − α, 2π + α, 3π − α, 4π + α and 5π − α.
Hence total six values in [0,5π] Ans.

ANSWERS FOR DO YOURSELF

π π
1. (i) (a) θ = nπ + (−1)n 6 , n ∈ I (b) θ = (2n + 1) 3 , n ∈ I

4nπ nπ
(c) θ= ,n ∈ I (d) θ= ,n ∈ I
3 2
π
(e) θ = nπ ± 12 , n ∈ I (f) θ = 2nπ + (−1)n+1 π, n ∈ I

π nπ kπ 3π
2: (i) (a) x = nπ + (−1)n+1 6 , n ∈ I (b) α= or α = + , n, k ∈ I
2 2 8

π
(c) θ = nπ ± 3 , n ∈ I

−1+√17 −1−√17
(d) θ = nπ + (−1)n α, where α = sin−1 ( ) or sin−1 ( ),n ∈ I
8 8
π −3π −π π 3π π
(ii) θ = {− 4 , , ,4, , 2}
4 2 4

π π
3: (i) (a) x = 2nπ − 4 , n ∈ I (b) 2 mπ + 2 , m ∈ I
mπ π nπ π
4: (i) θ = nπ or θ = ± 9 ; n, m ∈ I (ii) x= , n ∈ I and kπ ± 3 , k ∈ I
3 3
π
5: (i) D (ii) x= 4
2π 2π π 5π
𝟔: (i) ∪ [2nπ − , 2nπ + ] (ii) [6 , ]
n∈I 3 3 6

APNI KAKSHA 10
(MATHEMATICS) TRIGONOMETRIC EQUATION
EXERCISE 1

1. The number of solutions of the equation sin 2x − 2cos x + 4sin x = 4 in the interval [0, 5π] is-
(A) 6 (B) 4 (C) 3 (D) 5
2. Let A = {θ: sin (θ) = tan (θ)} and B = {θ: cos (θ) = 1} be two sets. Then -
(A) A = B (B) A ⊂ B and B − A ≠ ϕ
(C) A ⊄ B (D) B ⊄ A
3. The complete solution set of the inequality tan2 x − 2√2tan x + 1 ≤ 0 is -
π 3π π 3π
(A) nπ + 8 ≤ x ≤ 8 + nπ, n ∈ I (B) nπ + 4 ≤ x ≤ 4 + nπ, n ∈ I
π 3π π 2π
(C) nπ + 16 ≤ x ≤ + nπ, n ∈ I (D) nπ + 3 ≤ x ≤ + nπ, n ∈ I
8 3

4. The general solution of the equation tan2 α + 2√3tan α = 1 is given by-

nπ π
(A) α = 2 (n ∈ I) (B) α = (2n + 1) 2 (n ∈ I)
π nπ
(C) α = (6n + 1) 12 (n ∈ I) (D) α = (n ∈ I)
12

5. If 2tan2 θ = sec 2 θ, then the general solution of θ -

π π π π
(A) nπ + 4 (n ∈ I) (B) nπ − 4 (n ∈ I) (C) nπ ± 4 (n ∈ I) (D) 2nπ ± 4 (n ∈ I)
2
6. Number of principal solution(s) of the equation 4.16sin x
= 26sin x is
(A) 1 (B) 2 (C) 3 (D) 4
7. The general solution of equation 4cos2 x + 6sin2 x = 5 is-
π π
(A) x = nπ ± (n ∈ I) (B) x = nπ ± (n ∈ I)
2 4
3π
(C) x = nπ ± (n ∈ I) (D) None of these
2

8. If tan θ + tan 4θ + tan 7θ = tan θtan 4θtan 7θ, then θ =

nπ nπ nπ
(A) 4 (B) 7 (C) 12 (D) nπ
π π π
where n ∈ I, θ ≠ (2n + 1) 2 , θ ≠ (2n + 1) 8 , θ ≠ (2n + 1) 14
1−cos 2θ
9. If 1+cos 2θ = 3, then the general solution of θ is-
(A) 2nπ ± π/6 (B) nπ ± π/6 (C) 2nπ ± π/3 (D) nπ ± π/3
where n ∈ I
x
10. The number of solutions of the equation 2cos ( ) = 3x + 3−x is-
2
(A) 1 (B) 2 (C) 3 (D) None
11. The number of real solutions of the equation sin (ex ) = 5x + 5−x is-
(A) 0 (B) 1 (C) 2 (D) infinitely many
5π 5π
12. If x ∈ [− , ], then the greatest positive solution of 1 + sin4 x = cos 2 3x is-
2 2
5π
(A) π (B) 2π (C) (D) none of these
2

13. The general value of θ satisfying sin2 θ + sin θ = 2 is-

π π π π
(A) nπ + (−1)n 6 (B) 2nπ + 4 (C) nπ + (−1)n 2 (D) nπ + (−1)n 3
14. The number of solutions of the equation tan2 x − sec10 x + 1 = 0 in (0,10) is-
(A) 3 (B) 6 (C) 10 (D) 11

APNI KAKSHA 11
(MATHEMATICS) TRIGONOMETRIC EQUATION
15. The solution set of (5 + 4cos θ)(2cos θ + 1) = 0 in the interval [0,2π] is
π 2π π 2π 4π 2π 5π
(A) { , } (B) { , π} (C) { , } (D) { , }
3 3 3 3 3 3 3

16. The equation sin xcos x = 2 has :

(A) one solution (B) two solutions (C) infinite solutions (D) no solution
17. If tan2 θ − (1 + √3)tan θ + √3 = 0, then the general value of θ is :
π π π π
(A) nπ + 4 , nπ + 3 (B) nπ − 4 , nπ + 3
π π π π
(C) nπ + 4 , nπ − 3 (D) nπ − 4 , nπ − 3 where n ∈ I
18. If 0 ≤ x ≤ 3π, 0 ≤ y ≤ 3π and cos x. sin y = 1, then the possible number of values of the
ordered pair (x, y) is -
(A) 6 (B) 12 (C) 8 (D) 15
tan 2θ+tan θ
19. If 1−tan θtan 2θ = 0, then the general value of θ is
nπ nπ nπ
(A) nπ; n ∈ I (B) ;n ∈ I (C) (D) ;n ∈ I
3 4 6
where n ∈ I
20. The most general values of x for which sin x + cos x = mina∈R {1, a2 − 4a + 6} is given by
π
(A) 2nπ (B) 2nπ + 2
π π
(C) nπ + (−1)n 4 − 4 (D) None of these where n ∈ I

APNI KAKSHA 12
(MATHEMATICS) TRIGONOMETRIC EQUATION
EXERCISE 2 (JM)

1. The number of values of x in the interval [0,3π] satisfying the equation

2sin2 x + 5sin x − 3 = 0 is- [AIEEE 2006]
(1) 6 (2) 1 (3) 2 (4) 4
1
2. If 0 < x < π, and cos x + sin x = 2, then tan x is- [AIEEE 2006]
(1) (4 − √7)/3 (2) −(4 + √7)/3 (3) (1 + √7)/4 (4) (1 − √7)/4
3. Let A and B denote the statements
𝐀: cos α + cos β + cos γ = 0, B: sin α + sin β + sin γ = 0
3
If cos (β − γ) + cos (γ − α) + cos (α − β) = − 2, then - [AIEEE 2009]
(1) Both 𝐀 and 𝐁 are true (2) Both 𝐀 and 𝐁 are false
(3) 𝐀 is true and 𝐁 is false (4) 𝐀 is false and 𝐁 is true

4. The possible values of θ ∈ (0, π) such that sin (θ) + sin (4θ) + sin (7θ) = 0 are[AIEEE 2011]
2π π 4π π 3π 8π π 5π π 2π 3π 8π
(1) , , , , , (2) , , , , ,
9 4 9 2 4 9 4 12 2 3 4 9
2π π π 2π 3π 35π 2π π π 2π 3π 8π
(3) ,4,2, , , (4) ,4,2, , ,
9 3 4 36 9 3 4 9

5. If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation
cos x + cos 2x + cos 3x + cos 4x = 0, is:- [JEE(Main) 2016]
(1) 9 (2) 3 (3) 5 (4) 7
6. If 5(tan2 x − cos2 x) = 2cos 2x + 9, then the value of cos 4x is: [JEE (Main)-2017]
7 3 1 2
(1) − 9 (2) − 5 (3) 3 (4) 9
π x 1
7. If sum of all the solutions of the equation 8cos x ⋅ (cos ( 6 + x) cos (6 − x) − 2) = 1 in [0, π] is
kπ, then k is equal to [JEE (Main)-2018]
20 2 13 8
(1) 9 (2) 3 (3) 9 (4) 9
1
8. All the pairs (x, y) that satisfy the inequality 2√sin2 x − 2 sin x + 5 ⋅ 4 sin2 y ≤ 1 also satisfy the
equation: [JEE (Main)-2019]
(A) 2|sin x| = 3 sin y (B) 2 sin x = sin y
(C) sin x = 2 sin y (D) sin x = |sin y|
9. If α and β are the roots of equation (k + 1)tan x − √2λtan x = 1 − k and tan2 (α + β) = 50.
2

Find value of λ. [JEE (Main)-2020]

(A) 10 (B) 5 (C) 7 (D) 12
2 2 π
10. The number of solutions of the equation 32tan x
+ 32sec x
= 81,0 ≤ x ≤ 4 is:

[JEE (Main)-2021]
(A) 3 (B) 1 (C) 0 (D) 2
11. The number of solutions of |cos x| = sin x, such that −4π ≤ x ≤ 4π is: [JEE (Main)-2022]
(A) 4 (B) 6 (C) 8 (D) 12
12. If m and n respectively are the numbers of positive and negative value of θ in the interval
θ 9θ
[−π, π] that satisfy the equation cos 2θcos = cos 3θcos , then mn is equal to
2 2

APNI KAKSHA 13
(MATHEMATICS) TRIGONOMETRIC EQUATION
EXERCISE 3 (JA)

1. The number of integral values of k for which the equation 7cos x + 5sin x = 2k + 1 has a
solution is [JEE 2002 (Screening), 3]
(A) 4 (B) 8 (C) 10 (D) 12
𝟐. cos (α − β) = 1 and cos (α + β) = 1/e, where α, β ∈ [−π, π], numbers of pairs of α, β which
satisfy both the equations is [JEE 2005 (Screening)]
(A) 0 (B) 1 (C) 2 (D) 4
3. If 0 < θ < 2π, then the intervals of values of θ for which 2sin2 θ − 5sin θ + 2 > 0, is
[JEE-2006, 3]
π 5π π 5π π π 5π 41π
(A) (0, 6 ) ∪ ( 6 , 2π) (B) ( 8 , ) (C) (0, 8 ) ∪ ( 6 , ) (D) ( 48 , π)
6 6

4. The number of solutions of the pair of equations [JEE 2007, 3]

2sin2 θ − cos 2θ = 0
2cos 2 θ − 3sin θ = 0
in the interval [0,2π] is
(A) zero (B) one (C) two (D) four
−π π nπ
5. The number of values of θ in the interval ( 2 , 2 ) such that θ ≠ for n = 0, ±1, ±2
5

and tan θ = cot 5θ as well as sin 2θ = cos 4θ, is [JEE 2010, 3]

6. The positive integer value of n > 3 satisfying the equation [JEE 2011, 4]
1 1 1
π = 2π + 3π is
sin ( ) sin ( ) sin ( )
n n n

θ θ
7. Let θ, φ ∈ [0,2π] be such that 2 cos θ(1 − sin φ) = sin2 θ (tan 2 + cot 2) cos φ − 1,
√3
tan(2π − θ) > 0 and −1 < sin θ < − . Then φ cannot satisfy –
2
π π 4π 4π 3π 3π
(A) 0 < φ < (B) 2 < φ < (C) <φ< (D) < φ < 2π.
2 3 3 2 2

8. For x ∈ (0, π), the equation sin x + 2sin 2x − sin 3x = 3 has [JEE(Advanced)-2014, 3(-1)]
(A) infinitely many solutions (B) three solutions
(C) one solution (D) no solution
5
9. The number of distinct solutions of equation 4 cos 2 2x + cos 4 x + sin4 x + cos 6 x + sin6 x
= 2 in the interval [0,2π] is [JEE 2015, 4M,-0M]
10. Let a, b, c be three non-zero real numbers such that the equation
π π π
√3acos x + 2bsin x = c, x ∈ [− 2 , 2 ] has two distinct real roots α and β with α + β = 3. Then,
b
the value a is………. [JEE (Advanced)-2018, 3(0), P- 1]
11. Answer the following appropriately matching the list based on the information given in the
paragraph.
Let f(x) = sin (πcos x) and g(x) = cos (2πsin x) be two functions defined for x > 0. Define the
following sets whose elements are written in the increasing order.
X = {x: f(x) = 0}, Y = {x: f ′ (x) = 0}
[JEE (Advanced)-2019]
Z = {x: g(x) = 0}, W = {x: g ′ (x) = 0}

APNI KAKSHA 14
(MATHEMATICS) TRIGONOMETRIC EQUATION
List I contains the sets X, Y, Z and W. List II contains some information regarding these sets.
List I List II
π 3π
(I) X (P) ⊇ {2 , , 4π, 7π}
2
(II) Y (Q) an arithmetic progression
(III) Z (R) NOT an arithmetic progression
π 7π 13π
(IV) W (S) ⊇ { 6 , , }
6 6
π 2π
(T) ⊇ { 3 , , π}
3
π 3π
(U) ⊇ {6 , }
4
Which of the following is the only CORRECT combination ?
(A) (I), (Q), (U)
(B) (II), (Q), (T)
(C) (I), (P), (R)
(D) (II), (R), (S)
12. Consider the following lists:
List – I List - II
2π 2π
(I) {x ∈ [− , ] : cos x + sin x = 1} (P) has two elements
3 3

5π 5π
(II) {x ∈ [− 18 , 18 ] : √3tan 3x = 1} (Q) has three elements
6π 6π
(III) {x ∈ [− , ] : 2cos (2x) = √3} (R) has four elements
5 5
7π 7π
(IV) {x ∈ [− , ] : sin x − cos x = 1} (S) has five elements
4 4
(T) has six elements
The correct option is: [JEE (Advanced)-2022]

(A) (I) → (P); (II) → (S); (III) → (P); (IV) → (S)

(B) (I) → (P); (II) → (P); (III) → (T); (IV) → (R)
(C) (I) → (Q); (II) → (P); (III) → (T); (IV) → (S)
(D) (I) → (Q); (II) → (S); (III) → (P); (IV) → (R)

APNI KAKSHA 15
(MATHEMATICS) TRIGONOMETRIC EQUATION
ANSWER KEY
EXERCISE 1

1. (C) 2. (C) 3. (A) 4. (C) 5. (C) 6. (C) 7. (B)

8. (C) 9. (D) 10. (A) 11. (A) 12. (B) 13. (C) 14. (A)
15. (C) 16. (D) 17. (A) 18. (A) 19. (B) 20. (C)

EXERCISE 2 (JM)

1. 4 2. 2 3. 1 4. 1 5. 4 6. 1 7. 3

8. D 9. A 10. B 11. C 12. 25

EXERCISE 3 (JA)

1. B 2. D 3. A 4. C 5. 3 6. 7 7. A, C, D

8. D 9. 8 10. 0.50 11. B 12. B

APNI KAKSHA 16