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An equation involving one or more trigonometrical ratios of unknown angle is called a trigonometric equation e.g. cos2 x – 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)
· sin2 q = sin2 a , cos2 q = cos2 a, tan2q = tan2 aÞq = 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 1st or 4th 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
Þ ÞÞ
Þ sin2x = ± 1 Þ 2x = (2n + 1) Þ x = (2n+1) , n Î I
1: The general value of q satisfying both and is :
(A) 2np (B) 2np + 7p/6
(C) np + p/4 (D) 2np + p/4
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
(C) 3 (D) 4
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
(C) cot 4A (D) None of these
Solution: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A
= cot A
Hence (A) is the correct answer.
4: The value of sin 12°. sin48°.sin54° =
(A) 1/8 (B) 1/6
(C) 1/4 (D) 1/2
Solution: sin 12°. sin48°.sin54° =
Alternative Method
Let q = 12°
sin 12°. sin48°.sin54° =
tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°) is :
(A) 30° (B) 45°
(C) 60° (D) 90°
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°
6. The most general value of q satisfying 3 – 2cosq –4sinq –cos2q + sin2q =0:
(A) 2np (B) 2np + p/2
(C) 4np (D) 2np + p/4
Solution: 3 – 2cos q – 4 sin q – cos 2q + sin 2q = 0
Þ 3 – 2cos q – 4 sin q – 1 + 2sin2 q + 2sin q cos q = 0
Þ 2sin2q – 2cosq – 4sin q + 2sin q cos q + 2 = 0
Þ (sin2 q – 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
(C) 4 (D) 1
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
8: The minimum value of 3tan2q + 12 cot2q is:
(A) 6 (B) 8
(C) 10 (D) None of these
Solution: A.M. ³ G.M Þ (3tan2q +12 cot2q ) ³ 6
Þ 3 tan2q +12cot2q has minimum value 12.
9: If A + B + C = then the value of tanA + tanB + tanC is :
(A) 3 (B) 2
(C) > 3 (D) > 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.
10: Let 0 < A, B < satisfying the equalities 3 A + 2B = 1 and 3sin2A – 2sin2B = 0. Then A + 2B = :
(A) (B)
(C) (D) None of these.
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) 2a2/3 (B) a2/3
(C) 3a2/3 (D) 2a1/3
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 = 2a2/3.
12: If , then sin4a =
(A) a/2 (B) a
(C) a2/3 (D) 2a
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
13: If cos2q = and tan2 = tan2/3a, then cos2/3a + sin2/3a =
(A) 2a2/3 (B)
(C) (D) 2a1/3
Solution: cos2q = , tan2 = tan2/3a
tan3 = tan aÞ
= k
sin3 = k sin a ……(1)
cos3 = k cos a ……(2)
k2/3 sin2/3a + k2/3 cos a = 1
sin2/3a + cos2/3a =
Squaring and adding (1) and (2)
k2(sin2a + cos2a) = sin6 + cos6 =
k2 = 1 – sin2q = 1 – + cos2q
k2 = Þ k =
sin2/3a + cos2/3a = .
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 =
(C) (D)
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 = .
15: The value of is :
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
(C) 1 (D) 0
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.
17: The number of solutions of the equation x3 +2x2 +5x + 2cosx = 0 in [0, 2p] is:
(A) 0 (B) 1
(C) 2 (D) 3
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] .
18: The value of is equal to
(A) -1 (B)
Solution: .
19: sinnx=, where n is an odd natural number, then:
(A) = 1, = 2n (B) = 1, = n
(C) = 0, = n (D) = 0, = -n
Solution: sin nx = Im(ein x) = Im ((cosx + i sinx)n)
…..
Since n is odd, let n = 2 + 1
sin nx = – + ….
= – + + ….
20: If tanx = n. tany, n, then maximum value of (x – y) is equal to:
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,
21: If 3sinq + 5cosq = 5, then the value of 5sinq – 3cosq is equal to
(A) 5 (B) 3
(C) 4 (D) none of these
Solution: 3sinq = 5(1 – cosq) = 5 ´ 2sin2q/2 Þ tanq/2 = 3/5
5sinq – 3cosq = =
22: In a DABC, if cotA cotB cotC > 0, then the D is
(A) acute angled (B) right angled
(C) obtuse angled (D) does not exist
Solution: Since cotA cotB cotC > 0
cotA, cotB, cotC are positive ÞD is acute angled
23: If p < 2q < , then equals to
(A) –2cosq (B) –2sinq
(C) 2cosq (D) 2sinq
Solution: =
= 2 | sinq | = 2sinq as
24: If tanq = for some non-square natural number n, then sec2q is
(A) a rational number (B) an irrational number
(C) a positive number (D) none of these
where n is a non-square natural number so 1 – n ¹ 0.
Þ sec2q is a rational number.
25: The minimum value of cos(cosx) is
(A) 0 (B) –cos1
(C) cos1 (D) –1
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.
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
(C) 5/16 (D) none of these
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 = .
27: The maximum value of 4sin2 x + 3cos2x + is
(C) 9 (D) 4
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
28: If a and b are solutions of sin2 x + a sin x + b = 0 as well as that of cos2x + c cos x + d = 0, then sin(a + b) is equal to
Solution: According to the given condition, sina+sinb = –a and cosa +cosb= -c.
Þ Þ
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
(C) imaginary (D) greater than 1
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.
30: If then S equals
==
31: If in a DABC, ÐC =90°, then the maximum value of sin A sin B is
(A) (B) 1
(C) 2 (D) None
Solution: sinA sinB =
== = £
Þ Maximum value of sinA sinB =
32: If in a DABC, sin2A + sin2B + sin2C = 2, then the triangle is always
(A) isosceles triangle (B) right angled
(C) acute angled (D) obtuse angled
Solution: sin2 A + sin2 B + sin2C = 2 Þ 2 cos A cos B cos C = 0
Þ either A = 90o or B = 90o or C = 90o
33. Maximum value of the expression 2sinx + 4cosx + 3 is
(A) 2 + 3 (B) 2 - 3
(C) + 3 (D) none of these
Solution: Maximum value of 2sinx + 4cosx = 2.
Hence the maximum value of 2sinx + 4cosx +3 is
34: If sinq = 3sin(q + 2a), then the value of tan (q + a) + 2tana is
=3sin (q + a) cosa + 3cos (q + a) sina
35: If cos q = , then one of the values of tan is
(A) tan cot (B) tan cot
(C) sin sin (D) none of these
Solution: tan2 = =
= =
= tan2 cot2.
\ tan = ± tan cot .
36. If tan 2q. tan q = 1, then q is equal to
(A ) (B)
Solution: tan 2q . tan q = 1
.
37. If a is the root of 25 , then sin 2a is equal to
Solution: Since, a is the root of
38. The equation k possesses a solution if
(A) k > 6 (B)
(C) k > 2 (D) None of these.
Solution: We have k
But , therefore,
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
(C) x = (2n + 1) p, n Î Z (D) None of these.
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.
40. The equation is solvable if
Solution: We have
where
for y to be real.
Discriminant . . . (1)
But , therefore
. . . (2)
From (1) and (2), .
41. The set of values of x for which is
(A) f (B) p/4
but this value does not satisfy the given equation as and it reduces to indeterminate form.
42. If , then q is equal to
(A) p/3 (B) 2p/3
(C) p/6 (D) 5p/8
or
43. The value of the expression is
(A) 1/2 (B) 1
(C) 2 (D) None of these.
Solution: Given expression is
44. If , then q (only principal value) is
(C) 4p/3 (D) 5p/3
45. Number of solutions of in the interval [0, 2p] is
(A) 2 (B) 4
(C) 0 (D) None of these.
,
but
\ Solution does not exist.
46. If , then general solution for q is
47. Number of solutions of 11 sin x = x is
(A) 4 (B) 6
(C) 8 (D) None of these.
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.
48. If , then x is equal to
Solution: L.H.S.
and equality holds for
and R.H.S.
equality olds if .
Thus L.H.S. = R.H.S. for only.
49. General solution for q if , is
Solution: . . . (1)
and
(1) may holds true iff and both equal to 1 simultaneously. First common value of q is for which
and since periodicity of is p
and periodicity of is 2p, therefore, periodicity of is 2p. Therefore, general solution is .
50. If tan a and tan b are the roots of , then value of tan (a + b) is
Solution: are the roots of
and .
51. Number of solutions of the equation tan x = sec x = 2 cos x lying in the interval [0, 2p] is
Solution: The given equation can be written as
or –1
Hence, the required number of solutions is 2.
52. If tan mq + cot n q = 0, then the general value of q is
53. The general solution of the equation is
Solution: Let
Using these in the given equation, we get
or .
54. One solution of the equation is
Either sin q = 0 which gives q = n p
or which gives
Again
Thus, one solution of given equation is
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 +
(C) both
(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 -
(C) (D) None of these
Solution: Given, cosx + sinx =
Þcos x +sinx =
Þ cos
Þ x = 2np ± .
Þ x = 2np +, 2np - where n Î I.
57. The solution of the equation tan q . tan 2q = 1 is :
(A) np + (B) np -
(C) (D) np±
Solution: Given tan q. tan 2q = 1 Þ = 1
Þ 2 tan2q = 1 –tan2qÞ 3 tan2q = 1
Þ tan q = Þq = np±
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.
59. Solve for x, the equation sin3x + sin x cos x + cos3x = 1:
(A) 2mp (B) (4n + 1)
(C) Both (D) None of these
Solution: The given equation is sin3 x + cos3 x + sin x cos x = 1
Þ (sin x + cos x) (sin2 x – sin x cos x + cos2 x ) + 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)
60. The equation esinx – e–sinx – 4 = 0 has:
(A) no real solution (B) one real solution
(C) two real solutions (D) can't be determined
e2 sin x – 4esin x – 1 = 0 Þ esin 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.
61. If tan (p cos x) = cot (p sin x), then is
Solution: Given that tan (p cos x) = cos (p sin x)
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