# Inverse trigonometric functions & General Solution of Trigonometric Equations

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1 Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin θ = 8/17 then cos θ = 15/17 GE 15/17. c is rejected. b is the answer. OR Let sin = θ then sin θ = Cos θ = 1 sin θ = 1 = = 17 then G E =sin = = = = = 2. The value of Tan ( - ) = a) b) c) d) Ans: b. Solution: Let tan = θ then tan θ = Tan ( - tan ) = Tan ( - θ) = = =. 3. The domain of the function f(x) = is a) 1 x 2 b) -1 x 2 c) 2 x 4 d) 1 x 5 Ans: a. Solution : we know that sin is defined when - 1 x x 4 1 x

2 4. If cos ( 3 ) = tan ( ) then x = a) 0 b) 1 c) -1 d) 0,1, -1 Ans: d. Solution : 1) inspection method GE is satisfied by all values. Hence ans is d. or ii) cos = θ cos θ = x GE. cos3 θ = x 4 cos θ 3 cos θ = x 4x - 3x = x 4( x - x )= 0 x( x - 1) = 0 x = 0,1, If = + β and - β ) = then + = a) 1 + a b) a + b c) 1 + ab d) 1+ b Ans: c. Solution: α - β ) = 1 b α - β = cos 1 b ) = sin sin (α - β) = b Now sin α + cos β = sin α + 1 sin β 1 + sin α sin β = 1 + sin (α + β) sin (α - β) = 1 + a b If + then x satisfies a) 2x + 3x + 1 =0 b) 2x - 3x = 0 c) 2x + x - 1 =0 d) 2x + x + 1 =0 Ans: b Solution: ( x= 0 satisfies GE and b also. Hence ans b) Or sin + cos 1 sin sin + cos 1 - sin cos 1-2 sin = -2 θ (sin θ = x) 1 x = cos (-2 θ ) = cos 2 θ = 1 2 sin 2 θ 1 x = 1 2x x = 2x 2x - x = 0 x = 0 or x = x = 0 is the solution ( x = does not satisfies the equation )

3 7. If sec -1 x = cosec -1 y then sin -1 ( ) + sin-1 ( ) = a) Π b) π / 2 c) - π /2 d) - π Ans: b. Solution: Given sec -1 x = cosec -1 y cos ) = sin ) sin -1 ( ) + sin-1 ( ) = sin-1 ( ) + cos-1 ( ) =. 8. If tan ax tan bx = 0 ( a b) then the values of x form a series in a) AP b) G P c) H P d) AGP. Ans: a. Solution : tan ax = tan bx ax = n π + bx ax bx = n π x ( a b) = n π x =. put n = 1,2, x=, 9. If ( ) + ( ) = ( ) then x = etc. which are in AP. a) 0 b) c) d) π / 2. Ans: C Solution: w.k.t sin x )+ sin ( y ) = sin ( x1 + y 1 ) sin ( ) + sin ( ) = sin ( 1 )+ 1 ) ) = sin ( ( )/ 9 ) ) x =( )/ 9 Ans c 10. The value of = a) b) c) d). Ans: c Solution: GE = π + tan ( ) -. cot = π + tan ( ) - cot = π - tan ( ) - cot = π [tan ( ) + cot ] = π - = Ans c

4 11. Two angles of a triangle are and, then the third angle is a) b) c) d). Ans: b. Solution: A + B + C = Π where A = cot 2 = tan B= cot 3 = tan. A+ B = tan + tan = tan.. ] = tan 1 ] = C = π - ( A + B) = π - = a) :. : cos sin x cos cos 1 x 1 x sin cos x sin sin 1 x 1 x GE sin 1 x cos 1 x cos x sin x π 2. or Put x 0 then GE sin cos 0 cos sin sin 1 cos 1 π 2. sin If - = then the values of x are a) 0 or ½ b) 1 or 1/3 c) -1 or -1/3 d) -1,1/2 Ans: a Solution type 1: Inspection method: GE is satisfied by a) type 2: GE = cos x sin x = sin 1 x = sin x - sin x = sin 1 x sin 1 x = 2 sin x ( 1 x ) = sin 2 sin x ) = cos 2 sin x ) = 1 2 sin ( sin x) = 1 2x 1 x = 1 2x 2x x 2x x = 0 x ( 2x 1) = 0 x= 0, ½ ----

5 14. If + = then x = a) ± 3 b) 1 c) d) 0 Ans: b. Solution. Inspection method: ± 3 is not a solution since cos 3 is not defined. GE is satisfied by x = The value of + + = a) 0 b) cot 26 c) cot 1 d) cot 1 Ans: a. Solution: w.k.t cot a - cot b = cot cot 21 = cot. = cot 5 - cot 4 Similarly cot 13 = cot 4 - cot 3 And cot 8 = cot 3 - cot 5 GE = cot 5 - cot 4 + cot 4 - cot 3 + cot 3 - cot 5 = If +. = for 0< < then x = a) 1 b) 3) 0 d) Ans: a) Solution: clearly when x = 1 GE = ans: a) Or x = x = = x (2 x x ( 2 + x ) x = 0, 1. But x The value of 2 = a) b) c) d) Ans: b. searching method. put x = 1 then GE = 2tan 2-1] = 2 =. Put x = 1 in choices, Hence cot 1 = tan 1 = b Put x = -1 then GE = 2tan 2 - tan (π ) ] = = 2tan ] = - 2tan 2-1 ] = -2 ( ) = Hence answer must be either a or

6 Put x= -1 in a and b Then cot 1 = π = b tan 1 ) =. Hence answer is b If 2 =, then x is in a) [-1, 1 ] b) [-, 1] c) [- ] d) [ 0, 1]. Ans: c. Solution w.k. t. the range of RHS is [ 2 sin x sin x sin The solution set of the equation = 2 is a) { 1, 2} b) { -1, 2} c) {-1, 1, 0} d) {1, ½, 0} Ans: c) Solution : clearly is not defined. reject a) and b). consider d) = but =. = 2 hence reject d) Only possibility is ans c. or = 2 = x = x( ) = 2x x - = 0 x( 1 x ) = 0 x = 0, ±1 20. If in a triangle ABC, C is 90 0, then + a) b) c) Ans: c) d) = Solution: ABC is a right angled triangle. Put a= 3, b= 4 and c= 5. Then C= 90 Then tan + tan = tan + tan = tan + tan = = tan + tan = where m = 1. Ans: c.

7 21. The value of tan[ ) + )] = a) a b) c) b d) Ans: b). Solution Inspection method. Step1. Put a = 1, b= 0 Then GE = tan [tan 1)] = 1. When we put a=1, b= 0 in choices answer is either a or b Step2. a=0, b= 1 Then GE = tan [tan 1)+ tan 1) ] = tan [ ] = When we put a=0, b= 1 in choices Ans b) ( 1/a = 1/0 = ) If,,...a n is an AP with common difference d, then tan { } = a) b) c) Ans : b ) Solution: Given,,...a n are in AP - a 1 = a 3 a 2 =... a n - a n-1 = d d) Consider tan + tan tan = + tan tan = tan a - tan a + tan a - tan a tan a - tan a = tan a - tan a = tan = = tan =tan GE: tan tan ) = Hence answer is b.

8 23. If + = then x = a) 0 b) 1 c) 2-1 d) 1 Ans: d). Solution: Inspection method: Clearly x = o is not a solution since tan 0 =0 And cot 0 = LHS = When x= 1 LHS of GE = X = I is not a solution When x= 2-1, LHS of GE = is not a solution. x = -1 is only a solution If + + =, then X 1 x = a xyz b) 2xyz c) 2( x+ y + z) d) x y z Ans b Solution: It is a standard result. Or Inspection method. Step 1. put x= y = 1 and z = 0. Then sin x = π and choice b) and d) matches with this value. Step 2. Put x = and z = 1. Then sin x = π and choice b) matches with this value b is the answer If tan( x + y ) = 33 and x = then y = a) b) tan 30 c) tan d) Ans: d) tan Solution: x + y = tan 33 y =tan 33 - tan 3 = tan. =tan tan answer d)

9 26. 2 with x = a) sin 2x b) cos 2x c) cos x d) tan 1 x Ans: c Solution : put x = 1 then GE : When x = 1, a) and b) are not defined due to & c) = 0 and d) =. Hence Ans is c. 27. The general solution of sin5x = k where k is a real root of x - 1 = 0 is given by a) x = + 1 b x = + 1 (c) x = + 1 (d) x = n π + 1 Ans (a). Solution: Consider 2x - x +2x - 1 = 0 x (2x -1) +1 ( 2x 1) = 0 x +1 ) ( 2x 1) = 0 Since x is real x = ½. sin 5x = k = ½ = sin ( ) α = 5x = n π + 1 α = n 1 x = + 1 Ans (a) 28. If 5cos 2 θ = 0, θε (- π, π ),then θ = a) ( b), cos c) cos (d), π - cos Ans: (d ). Solution: GE = 5cos 2 θ + 2 cos + 1 = 0 5( 2 cos θ -1) + ( 1 + cos θ ) + 1 =0 10 x + x - 3 =0 where x = cos θ 10 x + 6x - 5x - 3 =0 2x ( 5x + 3 ) -1 ( 5x + 3 ) =0 x =, cos θ = ½ = cos, and cos θ = = cos( π - cos -1 (3/5)) Solution is, π - cos-1 (3/5)).

10 29. If The general solution of = 3 is given by θ = a) 2n π ± (b) n π ± (c). 2n π ± (b) n π ± Ans: (d) Solution: GE = = 3 tan θ = 3 tan θ =± 3 = tan( ± ) θ = n π ± If Sec x cos5x + 1 =0 where 0<x < 2 π, then x is equal to : a), b) c) d) none of these Ans: c Solution: GE : +1 = 0 cos5x + cosx = 0 2 cos 3x cos 2x =0 Cos 3x = 0 or cos2x =0 3x = or 2x = or X= or or or x = or inspection method. 31. The solution of the equation 2x = 1+ x is a) x = ( 2n + 1 ), n Є Z (b) x = n π, n Є Z.(C) x= ( 2n + 1 ), n Є Z (d) no solution. Ans: d Solution:GE: sin 2x = 1+ cos x Here cos x 0 The max value of LHS is 1 and minimum Vaue of RHS is 1. This is possible only when cos x = 0 and sin 2x = 1 x = and 2x = X =.This is not possible. Hence solution does not exists.

11 32. The most general value of θ satisfying the equation + tan θ - 1) 2 = 0 are given by a) n π ± b) n π + ( -1) n c). 2n π + (d) 2n π + Ans: c. Solution: a b = 0 a = 0 and b =0 1 2 sin θ = 0 and 3 tan θ - 1 = 0 sin θ = - ½ and tan θ = G.S. is θ = m π + ( -1) m and θ = mπ + only when m is odd m = 2n + 1 G.S. is θ = ( 2n + 1)π = 2n π The most general solution of θ satisfying Tan θ + tan( + θ ) = 2 is / are a) n π ±, n Є Z. b) 2n π +, n Є Z.c) 2n π ±, n Є Z d) ) 2n π +( -1) n, n Є Z. Both holds good Ans: a Solution: GE: Tan θ + tan( + θ ) = 2 Tan θ + tan 135 θ = 2 Tan θ + = 2 Tan θ + tan θ tan θ = tan θ tan θ = 3 tan θ = ± 3 θ = n π ±, n Є I The general solution of x for which cos2x,, and sin 2x are in Ap, are given by a) n π, n π + b) n π, n π + c) n π +, d) none of these. Ans: b. Solution: a,b,c are in AP a + c = 2b Cos2x + sin 2x = 1 Sin2x = 1 cos2x = 2 sin 2 x 2 sinx cos x - 2 sin 2 x = 0 2 sinx( cos x sinx) =0 sinx = o or cosx = sinx

12 sinx = 0 or tanx = 1 X = n π or x = n π +, n Є Z. 35. If sec θ + tan θ =1 then the general solution of θ= a) n π + b) 2n π + c) 2n π - d) 2n π ± Ans : c. GE : + = 1 cos θ sin θ = 2 cos( θ + ) = cos( 0 ) G.S. is θ + = 2 n π θ = 2 n π The equation sin cos has a) one solution b) two solutions c) infinite solutions d) no solution. Ans: d GE: sin3x = -2 < - 1. Hence solution does not exists The general solution of ( + ) = 2 is a) x = n π b) x = ( 4n + 1) c) x = ( 4n + 1) d) n π +. Ans: b. Solution clearly x = satisfies the GE If π x Π, π y Π and cos x + cosy = 2 then general solution is x = a) 2 n π + y b) 2 n π - y c) n π + y d) n π + ( -1) n y. Ans a. Solution : the Max. value of cosx is 1. Hence cos x + cosy = 2 cosx = 1 and cosy =1 x = 0 and y = 0 hence x y = 0 Hence cos( x y) = cos 0 X y = 2 n π x = 2 n π + y

13 39. The equation 3 x + 10cosx 6 =0 is satisfied if a) x = n π ± cos b) x = 2n π ± cos c) x = n π ± cos b) x = 2n π ± cos Ans: b. Solution GE.: 3( 1 - cos x ) + 10cosx 6 =0 3 3t + 10 t - 6 =0 where t = cosx 3t - 10 t +3 =0 3t - 9t - t +3 =0 3t( t 3 ) -1 ( t 3) =0 ( 3t -1) (t-3)=0 cosx = t = 3 > 1 No solution, Cosx = 1/3 = cos α where α = cos X = 2n π ± cos i.e. b). 40. If tan2x = tan, then the value of x = a) b) c) d) None of these Ans: a) Solution: GE: tan2x = tan 2x = n π + 2x - nπ x 2 = 0 X = 41. If the equation cos3x x + sin3x x= 0., then x = a) ( 2n+1) b) ( 2n - 1) c) d) n π,. Ans: a:) Solution: put n = 0 in answers a,b,c and d, Substitute x=0, x=, x=. GE is not satisfied when x =0 Hence c and d are not correct answers. GE is not satisfied when x = answer. Hence a is the answer. Or. Hence b is not the correct

14 cos3xcos x + sin3x sin x= 0. cos3x [ ( cos3x + 3 cosx )] + sin3x[ ( 3sinx - sin3x)] =0 ( cos 3x - sin 3x ) +3 ( cos3x cosx + sin3x sinx )} =0 {cos6x + 3 cos2x} =0 2x = A {cos3a + 3 cosa} =0 cos A0 cos 2x 0 cos2x = 0 2x = ( 2n + 1) x = ( 2n + 1) 42. If x and = 1, then all solutions of x are given by a) 2n π + b) ( 2n + 1) π - c) 2n π + 1 d) None of these. Ans : d. Solution: Since x, cos x 0,1,-1. Hence only possibility is sin x 3 sinx 2 = 0 sin x 2 sinx sinx 2 0 ( sinx - 1 ) ( sinx 2) =0 Sin x = 1 or 2 which is not possible as x and 2 > 1 Hence d is the answer. 43. If 1 + sin x + x + x +... = with 0<x < π and x, then x = a) b) c) Ans d. or d) Solution : put sinx = r then GE: 1 + r + r +... = ( in GP.) or With < 1 since 0<x < π and x. S = = = sinx = 1 sinx = x =

15 sinx = 1 - = = x = 60 or 120 Ans: d The general solution of of = cos2x 1 is a) 2n π b) c) n π d) (2n + 1) π Ans: c Solution: when n=1 a) π c) 2 π d is 3 π b) Among these GE is not satisfied by b) b is not the correct answer. Since GE is satisfied for π, 2π, 3 π also. General solution is θ = n π. Or tan x = cos2x 1 = - ( 1 cos2x) = - 2sin x sin x = - 2sin x cos x sin x ( 1 + 2cos x ) = 0 sin x = 0 sinx = 0 x= n π where n Є Z Cot θ =sin2 θ ( θ n π ),if θ = a), b), c) only d) only Ans b) Solution : when θ =, in GE LHS = RHS = 1. GE is satisfied by θ = also. and solution θ =,. Or GE = 2 sin θ cos θ cos θ = 2 sin θ cos θ cos θ ( 1-2 sin θ ) = 0 is not a Cos θ =0 θ = sin θ = ½ sin θ = ± θ = ±

16 46. If sin( cotθ ) =cos( tanθ) then the value of θ = a)n π + b ) 2n π ± c) n π - d) 2n π ± Ans: a) Solution: we know that sin = cos π cotθ = π tanθ cotθ = tanθ tan θ = 1 θ = n π If tan θ + 3 cot θ = 5 sec θ then θ = a) n π + ( -1) n, nє Z. b) n π + ( -1) n, nє Z c) n π + ( -1) n+ 1, nє Z or n π + ( -1) n, nє Z d) n π + ( -1) n, nє Z or n π + ( -1) n, nє Z Ans: b). Solution. GE: +3 sin θ + 3 cos θ = 5 sin θ cos θ = 5sin θ which is satisfied for x = only. answer is b) If tanx + tan4x + tan7x = tanx tan4x tan7x then x = a) n π / 3 b) n π / 4 c) n π / 6 d) n π / 12 Ans d). Solution : w. k.t if tan x + tany + tanz = tanx tany tanz then ] tan( x + y + z ) = 0. tan12x = 0 12x = n π x = n π / The general solution of sinx + sin7x = sin4x in ( 0, ) are a), b), c), d), Ans d) Solution: GE : sin7x + sin x sin4x =0 2 sin4x cos 3x sin 4x =0 sin4x( 2 cos3x - 1) =0 sin4x =0 4x = n π or x = n π / 4 = π / 4 in ( 0, ) cos 3x = = cos ( ) in ( 0, ) 3x = x = x =, Ans d

17 50. The number of solutions of cosx =, 0x 3 π is a) 3 b) 2 c) 4 d) 5 Ans: a. Solution: clearly 1 0 cosx = 1 + sinx ( = x if x 0 ) cos x sin x = 1 By inspection x = 0, 2π, 3 π /2 are 3 solutions in [0,3π] Ans a The set of values of x for which a) φ b) { n π +, n Є Z } c) Ans a. Solution: Clearly tan x = 1 = tan ( ) = 1 is d) { 2n π +, n Є Z } X = n π +, n Є Z. But for no values of x, sin2x is not defined, so that the equation has no solution. 52. The general solution of cos7 θ cos 5 θ = cos3θ cosθ is a), b), c), d ), Ans c. Solution : GE: cos7 θ cos 5 θ = cos3θ cosθ ( cos12 θ + cos2 θ ) = ( cos 4 θ + cos2 θ ) cos12 θ cos 4 θ = 0-2 sin 8θ sin 4θ =0 8θ = n or 4θ = n θ = Ans c If sec 2 x + + cosec 2 x = 4, then the value of x is a) b) c) d) Ans d. Clearly by inspection method d) is the answer ,

18 54. The equation 2 sin θ cos θ = x 2 + has a) one real solution b) no solution c) two real solutions d) θ = n π. ans b) Solution : GE : sin 2 θ = x 2 +. clearly sin 2 θ > 0. If x 1 solution does not exists. If 0<x< 1 then x < 1 but >1 hence x 2 + > 1 Solution does not exists. no solution 55. If A and B are acute angles such that sina = sin 2 B and 2 cos 2 A = 3 cos 2 B then A is d) a) b) c) Solution : Clearly by inspection method A=. Since when A=, B = from sina = sin 2 B. These values satisfies 2 cos 2 A = 3 cos 2 B also. a = is the ans Let n be a positive integer such that sin + cos =,then a) n = 4 b) n= 1,2,3,4,... c) n = 2 d) n = 6 Ans d. Solution: clearly by inspection n = 6. Or sin + cos squaring sin cos = sin = - 1 = which is true when n = 6. sin = n =6 only. 57. If sin 40 0 = k and cosx = 1 2 then the value of x in ( 0 0, ) are a) 40 o and 140 o b) 80 o and 280 o c) 40 o and 220 o d) 80 o and 260 o Ans: b. Solution : Now cosx = 1 2 sin = cos80 0. x = 80 0 and = x = 80 0, in ( 0 0, ) =

19 58. If 3 cos 2 x - 2 cos x sinx - 3 sin 2 x = 0 then x = a) + b) + c) d) n π + Ans: a) Solution : GE: 3 cos2x - 3 sin2x =0 3 cos2x = 3 sin2x = 3 tan2x = x = n π + x = The most general solution of + =0 a) 2n π +, n ЄZ, b) n π +, n ЄZ c) 2n π -, n ЄZ d) n π -, n ЄZ Ans: b. Solution: GE: log tanx = - log cotx log tanx = - log = log tanx log tanx = log tanx sinx = cosx Tanx = 1 x = n π +, n ЄZ ans :b The general solution of 3 tan( θ 15 o )= tan ( θ + 15 o ) is a) θ = ± b) n ± c) + ( -1) n d) 2n - Ans: c Solution: 3 tan( θ 15 o )= tan ( θ + 15 o ) 3 = 3 sin θ 15 cos θ 15 = sin θ 15 cos θ 15 2 sin θ 15 cos θ 15 + sin ( ) =0 sin2 θ + sin ( ) + sin ( ) =0 Sin 2 θ = 1 2 θ = n π + ( -1) n θ = ( -1) n

20 61. The most general solution of tan θ +1 = 0 & secθ - 2 =0 is a) 2n π - c) n π + ( -1) n Ans b) Solution: Given b) 2n π + d) 2n π + tan θ = quadrant take θ = 2 π = general solution is θ = 2n π + and cos θ = Ans b. θ in IV 62. If cosx + cosy = 1 and cosxcosy = ¼ then the general solutions are a) x = 2n π ±, y = 2k π ± b) 2n π ±, y = 2k π ± c) x = 2n π ±, y = 2k π ± b) n π ±, y = k π ± Ans: b. Solution: clearly by inspection method x =, y= is a solution. Hence ans is b). Or Cosx cosy = cosx cosy 4 cosxcosy = 1 4 =0 Now cosx + cosy = 1 and cosx - cosy = 0 cosx = ½ and cosy = ½ x = 2n π ±, y = 2k π ± ans : b = a b) - c) d) Ans : c Solution: put x= 0 then GE : In choices only a and c results in Hence answer is either a or c.. Put x = then GE: tan = tan 3= In choices a reduces to 15 0 and c reduces to Hence c is the answer.

21 Tan[ + cos-1 ( ) ] + Tan[ - cos-1 ( ) ] with x 0 a) 2x b) c) 2x d) 2 1 x Ans: b) Solution : put x = 1 in GE: = 2 reject d. Put x = in GE: tan( ) + tan ( ) = tan ( 750 ) + tan 15 0 = = 4 In choices b) gives 4. hence b is the correct answer If sin -1 ( ) + ) =, then x = a) 4 b) 5 c ) 1 d) 3 Ans: d. Solution: cosec ) = sin-1 ( ) GE : sin -1 ( ) + sin-1 ( ) = sin -1 ( ) = cos-1 ( ) clearly x = 3. sin -1 ( ) = cos-1 ( ) The value of cos( 2 + ) at x = 1/5 is a) b) c) d) 2 6 Ans: b Solution: cos( 2cos x + sin x ) = cos ( + cos x ) = - sin (cos x ) = - sin (sin 1 x ) = - 1 x At x = 1/5 ans is - 1 = - = The general solution of cosecx + cotx = is a) x = 2n π ± b) x = + - c) x = - d) x = 2n π ± - Ans: d. Solution: GE: cosecx + cotx = cosx = 3 sinx cosx + 3 sinx = - 1 cosx + cos( x + ) = cos G.S. is x = 2n π ± - ans :d..

22 68. If α, β and γ are the roots of the equation x 3 + m x 2 + 3x + m = 0, then the general solution of + + = a) ( 2n + 1) b) n π c) d) depend upon the value of x. Ans: b Solution : consider x 3 + m x 2 + 3x + m = 0 α = -b/a = -m, α = c/a = 3 ; α β γ = -d/a = -m Now =tan tan α + tan β + tan γ = tan = tan -1 (0) =0. G.S. is tan α +tan β +tan γ = n.where n Є Z. Ans b 69. The equation 3 cosx + 4 sinx = 6 has a) finite solution b) infinite solution c) one solution d) no solution. Ans: d Here r = > 1 where a= 3, b= 4 c = 6. Hence no solution. 70. The value of x satisfying the equation sinx + is given by a) 10 0 b) 30 0 c) 45 0 d) 60 0 = Ans: d Solution : Clearly by inspection x = 60 0 satisfies the equation. because sin = + = 71. If log 5 (1 + sinx) + log 5 (1 - sinx) = 0 then x in ( 0, ) is a) b) 0 c) d No value Ans d Solution: Clearly If log 5 [ (1 + sinx)( 1 sinx)] =0 1 sin 2 x = 0 cos 2 x =1 cosx = ± 1 x = 0, 0, ) no value.

23 72. If y = cosx and y = sin 5x then x = a) b) c) d) Ans: c Solution: clearly sin5x = cosx sin5x = sin ( - x) 5x = - x 6x = x = ans C 73. The value of ) ) + ) ) = a) Ans: d : b) c) = 2. d) 0 cos cos ) ) = cos cos 2π ) ) = sin sin ) = sin sin2π ) = - GE: = 0 Ans: d 74. If θ = )), then one of the possible value of θ is a) b) c) d) Ans: a) Solution: = θ = sin sin 600 = sin sin 120 = sin sin ) = sin sin 60 ) = 60 = 75. The value of sin( 2. = a) 0.96 b) 0.86 c) 0.94 d) Ans a) Solution: 0.8 = sin( 2 sin 0.8 = sin ( 2 θ ) where θ = sin = 2 sin θ cos θ = 2.. = = Ans a)

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