Page # 8 TRIGONOMETRIC EQUATION PH–II

EXERCISE – III HINTS & SOLUTIONS

Sol.4 sin2 = cos3

1 
Sol.1 (a) sin = = sin 
2 4 
 cos   2  = cos 3
2 

 = n + (–1)n , n 
4  
 3 = 2n    2 
2 

(b) tan (x – 1) = 3 = tan
3  
 = 2n + or  = 2n –
2 2

(x – 1) = n + , n 
3 2n  
 = + or  = 2n – , n 
5 10 2

x = n + + 1, n 
3  1 
=  2n   n, 
 2 5

(c) tan  = – 1 = tan  
 4  Sol.5 cot  = tan 8

    
 = n +   = n – , n   tan     = tan 8  8= n – 
 4  4  2  2

2  n 
3   9 = n +  = + ; n 
(d) cosec =  sin  = = sin 2 9 18
3 2 3
 1 
  =  n   ; n 
 = n + (–1)n , n   2 9
3
(e) 2 cot2 = cosec2
 cot2 = cosec2 – cot2 Sol.6 tan2 – (1 + 3 ) tan  + 3 =0
  (tan – 1) (tan – 3 )=0
 cot2 = 1 = cot2  = n ± ; n 
4 4
tan  = 1 or tan  = 3
Sol.2 sin 9 = sin   
9 = 2n +  or 9 = (2n + 1)  – , n   = n + , n  or  = n + ; n 
4 3
 8 = 2n or 10 = (2n + 1)n 
n  Sol.7  (0º, 90º)
 = or  = (2n + 1) ; n  sec2 . cosec2 + 2 cosec2 = 8
4 10
1  2 cos2 
Sol.3 cot + tan  = 2 cosec  =8 diff sin  0
sin2  cos2 
cos  sin  2 cos  0
 + =
sin  cos  sin   1 + 2 cos2  = 8 (1 – cos2) cos2
 8 cos4 – 6 cos2 + 1 = 0
cos2   sin2  2  (4 cos2 – 1) (2 cos2 – 1) = 0
 = If sin  0 & cos 0
sin  cos  sin  2
2
 sin – 2 sin cos  = 0 1  1 
 cos2  =   or cos2  =  
 sin  (1 – 2cos) = 0 2
   2
1  1
 but sin 0 so cos = = cos 1
2 3 cos = or cos =
2 2

 = 2n ± ; n   
3  = or =
3 4

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TRIGONOMETRIC EQUATION PH–II Page # 9

Sol.8 Sol.10 sin + sin 3 + sin 5 = 0

4 cos2 – 3 = 2 sin ( cos  0) sin 3 + 2sin 3 cos 2 = 0
 4 – 4 sin2 –3 = 2 sin  sin 3 (1 + 2 cos 2) = 0
 4sin2+ 2sin  – 1 = 0 sin 3 = 0 or 1 + 2 cos2 = 0

1  2 
2  4  4(4)(1)  1 5 3 = n cos2 = – = cos  
 sin  = = 2  3 
8 4
n  2 
5 1 ( 5  1) = , n  2 = 2m ±  
sin = or sin  = – 3  3 
4 4

  = m ± m 
  3
 sin = sin or sin  = –  cos 5 
10  
Sol.11 cos + sin  = cos 2 + sin 2
   3 
  = n + (–1)n or sin = sin   1 1
10  10   (cos + sin ) = (cos2 + sin2)
2 2

  3     
 = m + (–1)n   ,m
cos     = cos  2  
 10   4   4 

  
Sol.9 cot  – tan  = 2 2n ±     = 2 
4
1 – tan2 = 2 tan; tan   0  4
tan2 + 2 tan  – 1 = 0  
 2 – = 2n +  –  = 2n, n  I
2 44 4 4
tan = – =–1± 2
2  
& 2 – = 2n –  +
4 4
tan = 2 – 1 or tan  = – ( 2 + 1)
 2n 
 3 = 2n + q=  ,n  
2 3 6
 3
tan  = tan tan = – tan
8 8 Sol.12  (0, )
cos6 + cos4 + cos2 + 1 = 0
2cos5 cos + 2cos2 = 0
   3  2cos [cos5 + cos]
 = n + , n  or = tan  
8  8  cos = 0 or cos2 =0 or cos3 = 0
  
 1  3 = or  = or 3 = (2k + 1)
 =  2n   , n  or  = m + 2 4 2
 4  2 8
 3  5
= = or = ,
  4 4 6 6
= m – +
2 8
   3 5
= , , , ,
  6 4 2 4 6
 = (2m – 1) +
2 8
Sol.13 cos2x + cos22x + cos23x = 1
1 
 = ((2m – 1) + ) , m  1 cos 2x 1 cos 4 x 1 cos 6 x
4 2  + + =1
2 2 2
cos2x + cos4x + cos6x + 3 = 2
 1 
combined  =  n   n  cos2x + cos4x + cos6x + 1 = 0
 4  2 Previous Q.

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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 Page # 10 TRIGONOMETRIC EQUATION PH–II

   Sol.16 cosec = cot + 3

x = (2m + 1) , 2x = (2n + 1) , 3x = (2k + 1) , k 
2 2 2
 cosec 2 = cot2 + 2 3 cot + 3
 
m  x = (2n + 1) ,x = (2k + 1) , k   cosec2 – cot2 = 2 3 cot + 3
4 6
verticle line n 
2
 1 = 2 3 cot + 3  cot = –
2 3

 
 tan = – 3  tan = tan   
 3

 
 = n +   
 3
vertical line is induced
so remaining sol. 
 = n – , n 
 3
x = k ± k 
6 but IV quadrant is reject
sol. are Aliter :

  1 cos 
or x = (2m + 1) m , x = (2n + 1) ; = + 3 , sin  0
2 4 sin  sin 

  3 sin  + cos  = 1
n , x =  k   , k 
 6 
3 1 1
sin  + cos  =
2 2 2
Sol.14 sin2  – sin2 (n – 1)  = sin2 n is constant & n  0,1
 sin (n + n – ) sin (n – n + ) = sin2
 sin (2n – ) sin  = sin2
  1
cos     =
 sin = 0 or sin (2n – 1)  – sin = 0  3 2

 2n       2n       
 = m, m  or 2cos  2
 sin 
2
= 0
– = 2m ± , m 
    3 3
2 cos (n) sin (n – 1)  = 0
cos n = 0 or sin (n – 1) = 0 2
 = 2m or  = 2m + m 
3

n = (2k + 1) or (n – 1)  = p k, p,  reject cosec & cot is present. because does
2
not satisfy given equation (in IV quadrant)
( 2k  1)  p
= or  = ; p 
2 n (n  1) Sol.17 5sin + 2 cos = 5

 
Sol.15 10 tan 2(1  tan2 )
3 sin – cos  = 2 2 2
 + =5
2  2 
1 1 1  tan 1  tan

3
sin  – cos  = 2 2
2 2 2
  
  1   10 tan + 2 – 2tan2 = 5 + 5tan2
 sin     = = sin 2 2 2
 6  2 4
 
   7 tan2 – 10 tan +3=0
– = n + (–1)n 2 2
6 4
     
 = n +
 
+ (–1)n ; n    tan  1  7 tan  3  = 0
 2   2 
6 4

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TRIGONOMETRIC EQUATION PH–II Page # 11

     
 = n + or = n +   tan 3x = – 3 = tan     3x = n –
2 4 2  3 3

  = 2n + , n  n 
2 x= – , n 
3 9
3
or  = 2n + 2 ;  = tan–1 , n 
7 3 7
Sol.21 tan + sin  = ; tan2 + cos2 =
2 4
Sol.18 tan 2 tan  = 1
tan 2 = cot  7
tan2 + 1 – sin2 =
4
  
tan 2 =- tan     2 = n + –
2  2 3
tan2 – sin2 =
4
 
3 = n + 3 = n + 3
2 2
 (tan + sin ) (tan – sin ) =
4

 = (2n + 1) n 
6 3 3
 (tan – sin ) =
2n + 1  3k 2 4
Aliter : 1 – tan 2 tan  = 0
1
 tan 3 = not. defined  tan – sin  =
2
 
3 = n +  = (2n + 1) n  3
2 6 tan  + sin  =
2
2 tan  2 tan = 2
Alter: tan  = 1
1  tan2  tan  = 1
2 tan2 = 1 – tan2 
 = n + , n 
1  4
tan2 =  = n ±
3 6 2 sin  = 1
1
sin  =
Sol.19 tan + tan2 + 3 tan  tan 2 = 3 2

 tan + tan2 = 3 (1 – tan  tan 2) 

 = n + (–1)n , n 
6
tan   tan 2
 = 3 { 1 – tan2 tan  0}
(1  tan  tan 2) Sol.22 a tan  + bsec = c
 b sec = c – a tan 
 tan 3 = 3  3 = n =  b2 sec2 = c2 + a2 tan2 – 2 ac tan 
3
 b2 + b2 tan = c2 + a2 tan – 2ac tan
 1  tan 
 =  n   ; n 
 3 3  (a2 – b2) tan2 – 2ac tan  + c2 – b2 = 0
tan 

   2  2ac c2  b2
Sol.20 tan x . tan  x   ; tan x   = 3 tan + tan =
a2  b2
& tan . tan  =
a2  b2
 3   3 

     tan   tan  2ac / a 2  b 2

 tan x . tan   x  tan     3  x   = 3  tan ( + ) = =
1  tan  tan  1  (c 2  b 2 ) /(a 2  b2 )
3    

    2ac 2ac
 tanx tanx   x  tan   x  = – 3 = 2 2 2 2 =
3  3  a b c b a  c2
2

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
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 Page # 12 TRIGONOMETRIC EQUATION PH–II

Sol.23 a cos 2 + b sin 2 = c  4 (sinx + cosx) + (sin2x + 1) + 3 = 0

 b sin 2 = c – a cos2  4 (sinx + cosx) + (sinx + cos x)2 + 3 = 0
 b2 sin22 = c2 +a2cos22 – 2ac cos 2 Let sinx + cos x = t
 b2 – b2 cos22 = c2 + a2 cos22 – 2ac cos 2  4t + t2 + 3 = 0  t2 + 4t + 3 = 0
 (a2 + b2) cos22 – 2ac cos 2  (t + 1) (t + 3) = 0  sin x + cos x = –1
or sin x + cosx  –3 – 2 t 2
+ (c2 – b2) = 0
cos   1
 cos  x   = –
 4  2
2ac c 2  b2
cos2 + cos2 = & cos2 . cos2 =
a2  b 2 a2  b 2  3
x– = 2n ± n 
4 4
2ac
2cos2 – 1 + 2 cos2 – 1 =
a  b2
2
 x = 2n +  x = 2n –

n 
2
1  2ac  2  x = (2n + 1) n 
cos2 + cos2 =  
2  a 2  b2  But x  n

a2  ac  b2 Sol.26 2sinx = 3x2 + 2x + 3

= 2 2
a b In R.H.S. D = 4 – 36 = – 32
D 8
 min value of RHS = =
  4. 3 3
Sol.24 (sin2 + 3 cos 2)2 – 5 = cos  6  2  Max. value of L.H.S. = 2
L.H.S.  R.H.S.  No solution
2 vx R
  3 1   
 2 2 cos 2  2 sin 2 – 5 = cos   2  x 
    6 

 4 t2 – 5 = t  4t2 – t – 5 = 0 Sol.27 2 + 7 tan2 = 3.25 sec2 0º <  < 360º

325
   7(1 + tan2) – 5 = sec2
 (t + 1) (4t – 5) = 0 {let cos   2  = t 100
6 
13
 7 sec2 – 5 = sec2
 2  = –1 or cos   2   5
    4
 cos 
6 6 4
     28 sec2 – 20 = 13 sec2  15 sec2 = 20
   4
 –   2  = 2n +   2 = 2n + +  sec2 =
6  6 3

7 
= , n  [0, 2]  = 2n ±
12 6
 =30º, 150º, 210º, 330º
7 19
at n = 0  = & at n = 1   =
12 12
1 1
 log5 (sin x)  log15 cos x
Sol.28 51/2 + 52 = 152
x
sec 2
2  5 + 5 . 5log5 sin x = 3 5 = 15log15 cos x
Sol.25 1 + 2 cosecx = – (cosec x  0 x  n
2 sinx > 0 & cosx > 0
 x in Ist quadrant
sin x  2 1
 = 1 + sin x =
sin x x  3 cosx
2 cos 2
2
3 1 1
 (sinx + 2) (1 + cos x) + sin x = 0  cos x – sinx =
 2sinx + 2 cosx + sin x cosx + 2 = 0 2 2 2

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, email-info@motioniitjee.com
TRIGONOMETRIC EQUATION PH–II Page # 13

 2 cos4 cos 2 + (2cos22 – 1) = 0

    
 cos  x   = cos or x + = 2n ±  2cos4 cos2 + cos4 = 0
 6  3 6 3  cos4 [2cos2 + 1] = 0

  1
x = 2n + or x = 2n – (reject)  cos4 =0 or cos 2 = –
6 2 2

 2
In general, x = 2n + , n   4 = 2n or 2 = 2n ±
6 3

 3 5 7  2
Sol.29 sin + sin 5 = sin 3 0  = , , , = ,
8 8 8 8 3 3
2 sin 3 cos 2 = sin 3
sin3 (2cos2 – 1) = 0  3  5 2 7
= , , , , ,
1 8 8 3 8 3 8
sin 3 = 0 or cos =
2

 5 7 x x
sin 3 = 0 or 2 = , , Sol.32 2 + tanx . cot + cotx . tan =0
3 3 3 2 2

 2  5 7
 = 0, , ,   = , ,  [0, ] x x
3 3 6 6 6 tan x tan  2 x  2n
 2 + 2
x + tan x = 0
x 
  2 5 tan  , x  ( 2n  1)
 = 0, , , , & 2 2 2
6 3 3 6

x x
Sol.30 2 sin11x + cos3x + 3 sin3x = 0 2 tan tan x  tan2 x  tan2  x
tan  0
 2 2 =0  2
x  tan x 0
tan tan x
1 3 2
 cos3x + sin 3x = – sin11x
1x
2 2
2
 x x
 
 sin  3 x   = sin (–11x)   tan x  tan x  = 0  tanx = –tan
6  2 2


  x  x
 3x + = 2n + (–11x)  tanx = tan     x = n +   
6  2  2

 3x 2n
or 3x + = (2n + 1) – (–11)x  = n x =
6 2 3
  2
 14x = 2n – –8x + = 2n +  3
6 6

n  5
 x= – , n  –8x = 2n + x
7 84 6

n 5
x= – , n 
4 48
2
3
1 reject all sol. at 2n
Sol.31 cos . cos 2 . cos3 =
4
2
 2 cos2 . 2 cos cos 3 = 1 x = 2n ± , n 
 2cos2 [cos4 + cos2] = 1 3

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, email-info@motioniitjee.com
 Page # 14 TRIGONOMETRIC EQUATION PH–II

Sol.33 13  18 tan x = 6 tanx – 3 –2 < x < 2  

 cos 6x = cos   2 x 
 13 – 18 tan x = 9 (4tan2x
– 4 tan x + 1)  2 
 2
13 – 18 tanx = 36 tan x – 36 tanx + 9
 36 tan2x – 18 tanx – 4 = 0  
 cos6x = cos   2x 
 18 tan2x – 9 tanx – 2 = 0 2 
 (3 tanx – 2) (6 tanx + 1) = 0
 
1 2  6x = 2n ±   2x 
 tanx = – or tanx = 2 
6 3
 R.H.S. = (–ve)  
 6x = 2n + –2x or 6x = 2n – + 2x
Not possible  (LHS) = (+ve) 2 2

2   
tanx = = tan   8x = 2n + or 4x = 2n –
3 2 2
x = n +  n  n 
x = ,  + , – 2 + , –  +  x = + or x = –
4 16 2 8
2
where tan  =
3
  3  
min  , =
 16 8  16
1
Sol.34 cot 3 x  sin2 x  + 3 cos x  sin x  2
4 Sol.36 3 sin2A = sin2B
 3sin22A = sin22B
sin 3 x 2
= –  3 – 3 cos22A = 1 – cos22B
2 2
 ( 3 cos2A)2 – (cos2B)2 = 2
3 cosx + sinx  2
( 3 cos2A)2 – (cos2B)2 = 2
but 3 cosx + sinx  [–2, 2]
( 3 cos2A + cos2B) ( 3 cos2A – cos2B) = 2 ... (i)
 3 cosx + sinx = 2
3 1
& 3 sin2A + sin2B =
  2
cos  x   = 1
 6
 3 (1 – cos2A) + (1 – cos2B) = 3 –1

– 3 cos 2A – cos2B = –2

 
x– =0  x=
6 6  3 cos2A + cos 2B = 2 ....(ii)
check
by (i) & (ii)  3 cos2A – cos2B = 1 ...(iii)
1 1 1 1
0  + 22 = – 3
4 4 2 2 by (ii) & (iii) cos2A = 2A = 30º A = 15º
2

satisfy at x = 1
6 & cos2B = 2B = 60ºB = 30º
2

Sol.35 1 sin 2x – 2 cos3x = 0

 1 + sin2x = 2 cos23x
 1 + sin2x = 1 + cos 6x

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, email-info@motioniitjee.com