TRIGONOMETRIC EQUATIONS

An equation involving one or more trigonometrical ratios of unknown angle is called a trigonometric equation e.g. cos²x – 4 sin x = 1

It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle where as trigonometric equation is satisfied only for some values (finite or infinite) of unknown angle.

e.g. sec² x – tan² x = 1 is a trigonometrical identity as it is satisfied for every value of x Î R.

SOLUTION OF A TRIGONOMETRIC EQUATION

A value of the unknown angle which satisfies the given equation is called a solution of the equation e.g. sin q = ½ Þq = p/6 .

General Solution

Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

We use the following formulae for solving the trigonometric equations:

· sin q = 0 Þ q = np,

· cos q = 0 Þq = (2n + 1),

· tan q = 0 Þ q = np,

· sin q = sin a Þq = np + (–1)ⁿa, where aÎ [–p/2, p/2]

· cos q = cos aÞq = 2np ± a, where aÎ [ 0, p]

· tan q = tan a Þ q = np + a, where aÎ ( –p/2, p/2)

· sin²q = sin²a , cos²q = cos²a, tan²q = tan²aÞq = np±a,

· sin q = 1 Þq = (4n + 1),

· cos q = 1 Þ q = 2np ,

· cos q = –1 Þ q = (2n + 1)p,

· sin q = sin a and cos q = cos aÞ q = 2np + a.

Note:

· Everywhere in this chapter n is taken as an integer, If not stated otherwise.

· The general solution should be given unless the solution is required in a specified interval.

· a is taken as the principal value of the angle. Numerically least angle is called the principal value.

Method for finding principal value

Suppose we have to find the principal value of satisfying the equation sin = – .

Since sin is negative, will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical value of the angle will be greater than . If we approach it in clockwise direction the angle will be numerically less than . For principal value, we have to take numerically smallest angle.

So for principal value :

1. If the angle is in 1 st or 2nd quadrant we must select anticlockwise direction and if the angle if the angle is in 3rd or 4th quadrant, we must select clockwise direction.

2. Principal value is never numerically greater than .

3. Principal value always lies in the first circle (i.e. in first rotation)

On the above criteria will be or . Among these two has the least numerical value. Hence is the principal value of satisfying the equation sin = –.

Algorithm to find the principle argument:

Step 1: First draw a trigonometric circle and mark the quadrant, in which the angle may lie.

Step 2: Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for 3rd and 4th quadrants.

Step 3: Find the angle in the first rotation.

Step 4: Select the numerically least angle among these two values. The angle thus found will be the principal value.

Step 5: In case, two angles one with positive sign and the other with negative sign qualify for the numerically least angle, then it is the convention to select the angle with positive sign as principal value.

Example 1: Iftan = – 1, then will lie in 2nd or 4th quadrant.

For 2nd quadrant we will select anticlockwise and for 4th quadrant. we will select clockwise direction.

In the first circle two values and are obtained.

Among these two, is numerically least angle. Hence principal value is .

Example 2: If cos = , then will lie in 1^st or 4^th quadrant.

For 1st quadrant, we will select anticlockwise direction and for 4th quadrant, we will select clockwise direction.

In the first circle two values and are thus found.

Both and – have the same numerical value. In such case will be selected as principal value.

Illustration 17: Solve cot (sinx + 3) = 1.

Solution: sinx + 3 = Þ Þ n = 1 Þ sinx =

Þ x = or

Illustration 18: If sin 5x + sin 3x + sin x = 0, then find the value of x other than zero, lying between 0 £ x £.

Solution: sin 5x + sin 3x + sin x = 0 Þ (sin 5x + sin x) + sin 3x = 0

Þ 2 sin 3x cos 2x + sin 3x = 0 Þ sin 3x(2 cos 2x + 1) = 0

Þ sin 3x = 0; cos 2x = – Þ 3x = np, 2x = 2np±

The required value of x is .

Illustration 19: Find all acute angle a such that cos a cos 2a cos 4a = .

Solution: It is given that cosa cos2a cos4a =

Þ 2sina cosa cos2a cos4a = Þ 2sin2a cos2a cos4a =

Þ 2sin4a cos4a = sinaÞ sin8a – sina = 0

Þ 2sincos = 0

Either sin = 0 Þ Þa =

For n = 0 a = 0 which is not a solution.

Þa = n = 1, i.e. a =

or cos Þ = (2n + 1) Þa= (2n + 1) Þa =

Hence a = .

Illustration 20: Solve for x: .

Solution:

Þ ÞÞ

Þ sin2x = ± 1 Þ 2x = (2n + 1) Þ x = (2n+1) , n Î I

OBJECTIVE ASSIGNMENT

1: The general value of q satisfying both and is :

(A) 2np (B) 2np + 7p/6

Solution: Let us first find out q lying between 0 and 360°.

Since Þq = 210° or 330°

and Þq = 30° or 210°

Hence q = 210° or is the value satisfying both.

\The general value of

Hence (B) is the correct answer.

2: Ö3 cosec20° - sec20° =

(A) 1 (B) 2

Solution: Given =

Hence (D) is the correct answer.

3: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A =

(A) Cot A (B) tan 6A

Solution: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A

= tanA + 2tan2A + 4tan4A + 8

= cot A

Hence (A) is the correct answer.

4: The value of sin 12°. sin48°.sin54° =

(A) 1/8 (B) 1/6

Solution: sin 12°. sin48°.sin54° =

Alternative Method

Let q = 12°

sin 12°. sin48°.sin54° =

Hence (A) is the correct answer.

5: The smallest positive value of x (in degrees) for which

tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°) is :

(A) 30° (B) 45°

Solution: The relation may be written as

Þ Þ cos50°+ 2sin(2x + 50°) cos(2x + 50°) = 0

Þ cos50°+ sin (4x + 100°) = 0 Þ cos50° + cos(4x + 10°) = 0

Þ cos(2x + 30°) cos(2x – 20°) = 0 Þ x = 30°, 55°

Þ The smallest value of x = 30°

Hence (A) is the correct answer.

6. The most general value of q satisfying 3 – 2cosq –4sinq –cos2q + sin2q =0:

(A) 2np (B) 2np + p/2

Solution: 3 – 2cos q – 4 sin q – cos 2q + sin 2q = 0

Þ 3 – 2cos q – 4 sin q – 1 + 2sin²q + 2sin q cos q = 0

Þ 2sin²q – 2cosq – 4sin q + 2sin q cos q + 2 = 0

Þ (sin²q – 2sin q + 1) + cos q(sinq – 1) = 0

Þ (sin q – 1)[sin q – 1 + cos q] = 0

either sin q = 1

Þq = 2np + p/2 where n Î I

or, sin q + cos q =1

cos( q – p/4) = cos(p/4) Þq – p/4 = 2np±p/4

Þ q = 2np, 2np + p/2 where n Î I

Hence q = 2np, 2np + p/2.

Hence (A, B) is the correct answer.

7: If sinq = 3sin(q + 2a), then the value of tan (q + a) + 2tana is:

(A) 0 (B) 2

Solution: Given sin q = 3sin (q + 2a)

Þ sin (q + a-a) = 3sin (q + a + a)

Þ sin (q + a) cosa – cos(q + a) sina

= 3sin (q + a) cosa + 3cos (q + a) sina

Þ –2sin (q + a) cosa = 4cos (q + a) sina

Þ tan(q+a) + 2tana = 0

Hence (A) is the correct answer.

8: The minimum value of 3tan²q + 12 cot²q is:

(A) 6 (B) 8

Solution: A.M. ³ G.M Þ (3tan²q +12 cot²q ) ³ 6

Þ 3 tan²q +12cot²q has minimum value 12.

Hence (D) is the correct answer.

9: If A + B + C = then the value of tanA + tanB + tanC is :

(A) 3 (B) 2

Solution: tan(A + B) = tan( – C)

or, = tanC

or, tanA + tanB + tanC = tana tanB tanC

[since A.M. G.M.]

or, tanA tanB tanC

or, A B C 27 [cubing both sides]

or tanA tanB tanC 3

tanA + tanB + tanC 3.

Hence (A) is the correct answer.

10: Let 0 < A, B < satisfying the equalities 3 A + 2B = 1 and 3sin2A – 2sin2B = 0. Then A + 2B = :

(A) (B)

Solution: From the second equation, we have

sin2B = sin2A …(1)

and from the first equality

3A = 1 –2 B = cos2B …(2)

Now cos (A + 2B) = cosA. cos2B – sinA . sin2B

= 3 cosA . A – . sinA . sin2A

= 3cosA. A – 3A . cosA = 0

A + 2B = or

Given that 0 < A < and 0 < B < 0 < A + 2B < +

Hence A + 2B = .

Hence (C) is the correct answer.

11: If a cos³q + 3a cos q sin²q = x and a sin³q + 3a cos²q sin q = y, then (x + y)^2/3 + (x – y)^2/3 =

(A) 2a^2/3 (B) a^2/3

Solution: a cos³q + 3a cos q sin²q = x

a sin³q + 3a cos²q sin q = y

x + y = a[sin³q + cos³q + 3 sin q cos q(sin q + cos q)] = a(sinq + cosq)³

= sin q + cos q ……(1)

x – y = a[cos³q – sin³q + 3 cosq sin²q – 3 cos²q sin q] = a[cosq – sinq]³

= cos q – sin q ……(2)

(sin q + cos q)² + (cos q – sin q)² =

2 (sin²q + cos²q) =

(x + y)^2/3 + (x – y)^2/3 = 2a^2/3.

Hence (A) is the correct answer.

12: If , then sin4a =

(A) a/2 (B) a

Solution: Let a = sin 4qÞ = cos 2q + sin 2q and = cos 2q – sin 2q

(1 + ) tan a = (1 + )

Þ (1 + cos 2q + sin 2q) tan a = 1 + cos 2q – sin 2q

Þ = cot a

Þ = cot a Þ

Þ tan = tan Þq =

Þ a = sin 4q = sin (p – 4a) = sin 4 a

Hence (B) is the correct answer.

13: If cos²q = and tan² = tan^2/3a, then cos^2/3a + sin^2/3a =

(A) 2a^2/3 (B)

Solution: cos²q = , tan² = tan^2/3a

tan³ = tan aÞ

= k

sin³ = k sin a ……(1)

cos³ = k cos a ……(2)

k^2/3 sin^2/3a + k^2/3 cos a = 1

sin^2/3a + cos^2/3a =

Squaring and adding (1) and (2)

k²(sin²a + cos²a) = sin⁶ + cos⁶ =

k² = 1 – sin²q = 1 – + cos²q

k² = Þ k =

sin^2/3a + cos^2/3a = .

Hence (B) is the correct answer.

14: If 3 sin²a + 2 sin²b = 1 and 3 sin 2a –2 sin 2b = 0, where a, b are positive acute angles, then a + 2b =

(A) (B)

Solution: 3 sin²a + 2 sin²b = 1 ……(1)

3 sin 2a = 2 sin 2b ……(2)

3 sin²a = 1 – 2 sin²b = cos 2b

3 sin a sin a = cos 2b ……(3)

from equation (2)

3 . 2 sin a cos a = 2 sin 2b

3 sin a =

from equation (3)

sin a = cos 2b

cos a cos 2b – sin a sin 2b = 0

cos (a + 2b) = 0

Þa + 2b = .

Hence (A) is the correct answer.

15: The value of is :

(A) (B)

Solution:

Hence (A) is the correct answer.

16: The number of solutions of sin³ x cos x + sin² x cos² x + sin x cos³ x = 1 in [0, 2p] is

(A) 4 (B) 2

Solution: sin x cos x [sin²x + sin x cos x + cos² x] = 1

Þ sin x cos x + (sin x cos x)² = 1

sin² 2x + 2 sin 2x –4 = 0 Þ sin 2x = , which is not possible.

Hence (D) is the correct answer.

17: The number of solutions of the equation x³ +2x² +5x + 2cosx = 0 in
[0, 2p] is:

(A) 0 (B) 1

Solution: Let f(x) = x³ + 2x² + 5x +2 cosx

Þ f¢(x) = 3x² +4x + 5 – 2 sinx

= 3

Now "x ( as -1 £ sinx £ 1)

Þ f¢(x) > 0 " x

Þ f(x) is an increasing function.

Now f(0) = 2

Þ f(x) = 0 has no solution in [ 0, 2p] .

Hence (A) is the correct answer.

18: The value of is equal to

(A) -1 (B)

Solution: .

Hence (D) is the correct answer.

19: sinnx=, where n is an odd natural number, then:

(A) = 1, = 2n (B) = 1, = n

Solution: sin nx = Im(e^{in x}) = Im ((cosx + i sinx)ⁿ)

…..

Since n is odd, let n = 2 + 1

sin nx = – + ….

= – + + ….

Hence (C) is the correct answer.

20: If tanx = n. tany, n, then maximum value of (x – y) is equal to:

(A) (B)

Solution: tanx = n tany, cos(x – y)

= cosx. cosy + sinx.siny.

cos(x – y) = cosx.cosy(1 + tanx.tany)

= cosx. cosy (1 + n tan²y)

Now,

Hence (D) is the correct answer.

21: If 3sinq + 5cosq = 5, then the value of 5sinq – 3cosq is equal to

(A) 5 (B) 3

Solution: 3sinq = 5(1 – cosq) = 5 ´ 2sin²q/2 Þ tanq/2 = 3/5

5sinq – 3cosq = =

Hence (B) is the correct answer.

22: In a DABC, if cotA cotB cotC > 0, then the D is

(A) acute angled (B) right angled

Solution: Since cotA cotB cotC > 0

cotA, cotB, cotC are positive ÞD is acute angled

Hence (A) is the correct answer.

23: If p < 2q < , then equals to

(A) –2cosq (B) –2sinq

Solution: =

= 2 | sinq | = 2sinq as

Hence (D) is the correct answer.

24: If tanq = for some non-square natural number n, then sec2q is

(A) a rational number (B) an irrational number

Solution:

where n is a non-square natural number so 1 – n ¹ 0.

Þ sec2q is a rational number.

Hence (A) is the correct answer.

25: The minimum value of cos(cosx) is

(A) 0 (B) –cos1

Solution: cos x varies from –1 to 1 for all real x.

Thus cos(cosx) varies from cos1 to cos0 Þ minimum value of cos(cosx) is cos1.

Hence (C) is the correct answer.

26: If sin x cos y = 1/4 and 3 tan x = 4 tan y, then find the value of sin (x + y).

(A) 1/16 (B) 7/16

Solution: 3 tan x = 4 tan y Þ 3 sin x cos y = 4 cos x sin y

Þ 3/4 = 4 cos x sin y Þ cos x sin y = 3/16

\ sin (x + y) = sin x cos y + cos x sin y = .

Hence (B) is the correct answer.

27: The maximum value of 4sin² x + 3cos²x + is

(A) (B)

Solution: Maximum value of 4sin²x + 3cos²x i.e. sin²x + 3 is 4 and that of sin+ cos is = , both attained at x = p/2. Hence the given function has maximum value

Hence (A) is the correct answer.

28: If a and b are solutions of sin²x + a sin x + b = 0 as well as that of cos²x + c cos x + d = 0, then sin(a + b) is equal to

(A) (B)

Solution: According to the given condition, sina+sinb = –a and cosa +cosb= -c.

Þ Þ

Hence (D) is the correct answer.

29: If sina, sinb and cosa are in G.P, then roots of the equation x² + 2x cot b+ 1 = 0 are always.

(A) equal (B) real

Solution: sina, sinb, cosa are in G.P.

Þ sin²b = sina cosaÞ cos2b = 1 – sin2b ³ 0

Now, the discriminant of the given equation is

4cot²b – 4 = 4 cos2b× cosec²b³ 0 Þ Roots are always real.

Hence (B) is the correct answer.

30: If then S equals

(A) (B)

Solution:

Hence (C) is the correct answer.

31: If in a DABC, ÐC =90°, then the maximum value of sin A sin B is

(A) (B) 1

Solution: sinA sinB =

== = £

Þ Maximum value of sinA sinB =

Hence (A) is the correct answer.

32: If in a DABC, sin²A + sin²B + sin²C = 2, then the triangle is always

(A) isosceles triangle (B) right angled

Solution: sin² A + sin² B + sin²C = 2 Þ 2 cos A cos B cos C = 0

Þ either A = 90^o or B = 90^o or C = 90^o

Hence (B) is the correct answer.

33. Maximum value of the expression 2sinx + 4cosx + 3 is

(A) 2 + 3 (B) 2 - 3

Solution: Maximum value of 2sinx + 4cosx = 2.

Hence the maximum value of 2sinx + 4cosx +3 is

Hence (A) is the correct answer.

34: If sinq = 3sin(q + 2a), then the value of tan (q + a) + 2tana is

(A) 3 (B) 2

Solution: Given sin q = 3sin (q + 2a)

Þ sin (q + a-a) = 3sin (q + a + a)

Þ sin (q + a) cosa – cos(q + a) sina

=3sin (q + a) cosa + 3cos (q + a) sina

Þ –2sin (q + a) cosa = 4cos (q + a) sina

Þ tan(q+a) + 2tana = 0

Hence (D) is the correct answer.

35: If cos q = , then one of the values of tan is

(A) tan cot (B) tan cot

Solution: tan² = =

= =

= tan² cot².

\ tan = ± tan cot .

Hence (A) is the correct answer.

36. If tan 2q. tan q = 1, then q is equal to

(A ) (B)

Solution: tan 2q . tan q = 1

Hence (B) is the correct answer.

37. If a is the root of 25 , then sin 2a is equal to

(A) (B)

Solution: Since, a is the root of

Hence (B) is the correct answer.

38. The equation k possesses a solution if

(A) k > 6 (B)

Solution: We have k

But , therefore,

Now,

Hence (B) is the correct answer.

39. The general solution of the equation tan 3x = tan 5x is

(A) x = np/2, n Î Z (B) x = np, n Î Z

Solution: We have tan 3x = tan 5x

if n is odd, then x = np/2, gives the extraneous solutions. Thus, the solution of the given equation will be given by x = np/2, where n is even say n = 2 m, m Î Z. Hence, the required solution is x = m p, m Î Z.

Hence (B) is the correct answer.

40. The equation is solvable if

(A) (B)

Solution: We have

where

for y to be real.

Discriminant . . . (1)

But , therefore

. . . (2)

From (1) and (2), .

Hence (B) is the correct answer.

41. The set of values of x for which is

(A) f (B) p/4

Solution:

but this value does not satisfy the given equation as and it reduces to indeterminate form.

Hence (A) is the correct answer.

42. If , then q is equal to

(A) p/3 (B) 2p/3

Solution:

Hence (C) is the correct answer.

43. The value of the expression is

(A) 1/2 (B) 1

Solution: Given expression is

Hence (B) is the correct answer.

44. If , then q (only principal value) is

(A) p/3 (B) 2p/3

Solution: .

Hence (A) is the correct answer.

45. Number of solutions of in the interval [0, 2p] is

(A) 2 (B) 4

Solution:

but

\ Solution does not exist.

Hence (C) is the correct answer.

46. If , then general solution for q is

(A) (B)

Solution:

Hence (B) is the correct answer.

47. Number of solutions of 11 sin x = x is

(A) 4 (B) 6

Solution: 11 sin x = x . . . (1)

On replacing n by –, we have 11 sin (–x) = –x

So for every positive solution, we have negative solution also and x = 0 is satisfying (1), so number of solution will always be odd. Therefore, (d0 is appropriate choice.

Hence (D) is the correct answer.

48. If , then x is equal to

(A) (B)

Solution: L.H.S.

and equality holds for

and R.H.S.

equality olds if .

Thus L.H.S. = R.H.S. for only.

Hence (B) is the correct answer.

49. General solution for q if , is

(A) (B)

Solution: . . . (1)

and

(1) may holds true iff and both equal to 1 simultaneously. First common value of q is for which

and

and since periodicity of is p

and periodicity of is 2p, therefore, periodicity of is 2p. Therefore, general solution is .

Hence (A) is the correct answer.

50. If tan a and tan b are the roots of , then value of tan (a + b) is

(A) (B) 1

Solution: are the roots of

and .

Hence (C) is the correct answer.

51. Number of solutions of the equation tan x = sec x = 2 cos x lying in the interval [0, 2p] is

(A) 0 (B) 1

Solution: The given equation can be written as

or –1

Hence, the required number of solutions is 2.

Hence (C) is the correct answer.

52. If tan mq + cot n q = 0, then the general value of q is

(A) (B)

Solution: The given equation can be written as

Hence (A) is the correct answer.

53. The general solution of the equation is

(A) (B)

Solution: Let

and

Using these in the given equation, we get

or .

Hence (D) is the correct answer.

54. One solution of the equation is

(A) (B)

Solution: The given equation can be written as

Either sin q = 0 which gives q = n p

or which gives

Now,

Again

Thus, one solution of given equation is

Hence (A) is the correct answer.

55. Solve for x and y, the equations:

xy + 3x cosy. y = 14

xy + 3x. y siny = 13

(A) y = where 2n < y < 2n +

(B) y = where 2n + < y < 2n +

(D) None of these

Solution: Clearly, x 0 dividing the equations, we get

by componendo and dividenodo, we get

or, = 27 =

or, =

dividing numerator and denominator by cosy, we get

or, .

siny = , cosy = (when y is in 1st quadrant)

and siny = - and cosy = - (when y is in 3rd quadrant)

When y is in first quadrant.

When y is in 3rd quadrant.

Hence y = where 2n < y < 2n +

and y = where 2n + < y < 2n +

56. The solution of sinx + cosx = is :

(A) 2np + (B) 2np -

Solution: Given, cosx + sinx =

Þcos x +sinx =

Þ cos

Þ x = 2np ± .

Þ x = 2np +, 2np - where n Î I.

Hence (A, B) is the correct answer.

57. The solution of the equation tan q . tan 2q = 1 is :

(A) np + (B) np -

Solution: Given tan q. tan 2q = 1 Þ = 1

Þ 2 tan²q = 1 –tan²qÞ 3 tan²q = 1

Þ tan q = Þq = np±

Hence (D) is the correct answer.

58. Find the general solution of the equation

sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x:

(A) (B) np -

Solution: Given sin x – 3 sin 2x + sin 3x = cos x –3 cos 2x + cos 3x

Þ 2 sin 2x cos x – 3 sin 2x = 2 cos x cos 2x – 3 cos 2x

Þ sin 2x (2 cos x –3) = cos 2x (2 cos x –3) Þ sin 2x = cos 2x

( cos x ¹ 3/2)

Þ tan 2x = 1 Þ 2x = np + Þ x = , n Î I.

Hence (C) is the correct answer.

59. Solve for x, the equation sin³x + sin x cos x + cos³x = 1:

(A) 2mp (B) (4n + 1)

Solution: The given equation is sin³x + cos³ x + sin x cos x = 1

Þ (sin x + cos x) (sin²x – sin x cos x + cos²x ) + sin x cos x – 1 = 0

Þ (1 – sin x cos x)[sin x + cos x – 1] = 0

Either 1 – sin x cos x = 0 Þ sin 2 x = 2 which is not possible

Or, sin x + cos x – 1 = 0 Þ cos (x – p/4) = Þ ±

Þ x = 2mp and x = (4n + 1)

Hence (C) is the correct answer.

60. The equation e^sinx – e^–sinx – 4 = 0 has:

(A) no real solution (B) one real solution

Solution: The given equation can be written as

e^{2 sin x} – 4e^{sin x} – 1 = 0 Þ e^{sin x} = = 2 +

Þ sin x = ln (2 + ) (ln (2 – ) not defined as (2 – ) is negative)

Now, 2 + > e Þ ln (2 + ) > 1 Þ sin x > 1

Which is not possible. Hence no real solution.

Hence (A) is the correct answer.

61. If tan (p cos x) = cot (p sin x), then is

(A) (B)

Solution: Given that tan (p cos x) = cos (p sin x)

Hence (B) is the correct answer.

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