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```TRIGONOMETRIC EQUATIONS

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

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

(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

= 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

(C)        1/4                                             (D)        1/2

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°

(C)        60°                                            (D)        90°

Solution:              The relation may be written as

Þ

Þ

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

(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

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

(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.

Hence (D) is the correct answer.

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.

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)

(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.

Hence (A) is the correct answer.

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

Hence (B) is the correct answer.

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 = .

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)

(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 = .

Hence (A) is the correct answer.

15:                     The value of  is :

(A)                      (B)

(C)                      (D)

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

(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.

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

(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] .

Hence (A) is the correct answer.

18:                     The value of  is equal to

(A)        -1                                              (B)

(C)                                               (D)

Solution:              .

Hence (D) is the correct answer.

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 = – + ….

= – + + ….

=

Hence (C) is the correct answer.

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

(A)                                          (B)

(C)                                          (D)

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

(C)  4                                                  (D) none of  these

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

(C)  obtuse angled                           (D)  does  not  exist

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

(C)  2cosq                                          (D)  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

(C)  a positive number                     (D)  none of these

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

(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.

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

(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 = .

Hence (B) is the correct answer.

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

(A)                                            (B)

(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

Hence (A) is the  correct answer.

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

(A)                                         (B)

(C)                                         (D)

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

Þ

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

(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.

Hence (B) is the  correct answer.

30:                   If  then S equals

(A)                                        (B)

(C)                                        (D)

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

(C) 2                                                   (D) None

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

(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

Hence (B) is the correct answer.

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

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

(C) 1   (D) 0

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

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

(C) sin  sin                                      (D) none of these

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)

(C)                                          (D)        None of these.

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)

(C)                                                              (D)

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)

(C)        k > 2                                          (D)        None of these.

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

(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.

Hence (B) is the correct answer.

40.                     The equation is solvable if

(A)                               (B)

(C)                                       (D)        None of these.

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

(C)                (D)

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

(C)        p/6                                            (D)        5p/8

Solution:

or

.

Hence (C) is the correct answer.

43.                     The value of the expression is

(A)        1/2                                             (B)         1

(C)        2                                               (D)        None of these.

Solution:              Given expression is

.

Hence (B) is the correct answer.

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

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

(C)        4p/3                                           (D)        5p/3

Solution:              .

Hence (A) is the correct answer.

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

(A)        2                                               (B)         4

(C)        0                                               (D)        None of these.

Solution:

,

but

\   Solution does not exist.

Hence (C) is the correct answer.

46.                    If , then general solution for q is

(A)                                          (B)

(C)                                   (D)        None of these.

Solution:

.

Hence (B) is the correct answer.

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.

Hence (D) is the correct answer.

48.                     If , then x is equal to

(A)                                                 (B)

(C)                                                    (D)        None of these.

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)

(C)                                       (D)        None of these.

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

(C)                                                    (D)        None of these.

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

(C)        2                                               (D)        3

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)

(C)                                             (D)

Solution:              The given equation can be written as

or

.

Hence (A) is the correct answer.

53.                     The general solution of the equation is

(A)                          (B)

(C)                          (D)

Solution:              Let

or

and

or

Using these in the given equation, we get

or

or   .

Hence (D) is the correct answer.

54.                     One solution of the equation is

(A)                    (B)

(C)                                   (D)        None of these.

Solution:              The given equation can be written as

or

or

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 +

(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

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 -

(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±

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 -

(C)                                                                    (D)        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)

(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)

Hence (C) is the correct answer.

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

Solution:              The given equation can be written as

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.

Hence (A) is the correct answer.

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

(A)                                                 (B)

(C)        0                                               (D)        None of these.

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

or

.

Hence (B) is the correct answer.

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