Question-1 :-  Find the principal and general solutions of tan x = √3.

  We know that tan π/3 = √3 and
  tan(4π/3) = tan(π + π/3) = tan π/3 = √3
  Therefore, principal solutions are π/3 and 4π/3.

  Now, tan x = tan π/3
  x = nπ + π/3, where  n ∈ Z.
     

Question-2 :-  Find the principal and general solutions of sec x = 2.

  We know that sec π/3 = 2 and
  sec(5π/3) = sec(2π - π/3) = sec π/3 =2
  Therefore, principal solutions are π/3 and 5π/3.

  Now, sec x = sec π/3
       cos x = cos π/3          [sec x = 1/cos x]
  x = 2nπ ± π/3, where  n ∈ Z.
   

Question-3 :-  Find the principal and general solutions of cot x = -√3.

  We know that cot π/6 = √3
  cot(5π/6) = cot(π - π/6) = -cot π/6 = -√3
  cot(11π/6) = cot(2π - π/6) = -cot π/6 = -√3
  Therefore, principal solutions are 5π/6 and 11π/6.

  Now, cot x = cot(5π/6)
       tan x = tan(5π/6)        [cot x = 1/tan x]
  x = nπ + 5π/6, where  n ∈ Z.
    

Question-4 :-  Find the principal and general solutions of cosec x = – 2.

  We know that cosec π/6 = 2 
  cosec 7π/6 = cosec(π + π/6) = -cosec π/6 = -2
  cosec 11π/6 = cosec(2π - π/6) = -cosec π/6 = -2
  Therefore, principal solutions are 7π/6 and 11π/6.

  Now, cosec x = cosec(7π/6)
       sin x = sin(7π/6)        [cosec x = 1/sin x]
  x = nπ + (-1)ⁿ 7π/6, where  n ∈ Z.
     

Question-5 :-  Find the general solutions of cos 4x = cos 2x.

  cos 4x = cos 2x
  cos 4x - cos 2x = 0
  -2 sin 6x/2 sin 2x/2 = 0   [cos a - cos b = -2 sin(a + b)/2 . sin(a - b)/2]
  sin 3x sin x = 0
  sin 3x = 0 or sin x = 0
  3x = nπ or x = nπ, where  n ∈ Z.
   x = nπ or x = nπ, where  n ∈ Z.
   

Question-6 :-  Find the general solutions of cos 3x + cos x – cos 2x = 0.

  cos 3x + cos x – cos 2x = 0
  2 cos 4x/2 cos 2x/2 - cos 2x = 0   [cos a + cos b = 2 cos(a + b)/2 . cos(a - b)/2]
  2 cos 2x cos x - cos 2x = 0
  cos 2x(2 cos x - 1) = 0
  cos 2x = 0 or 2 cos x - 1 = 0
  cos 2x = 0 or cos x = 1/2
  2x = π/2 or x = π/3
  2x = (2n + 1)π/2 or x = 2nπ ± π/3, where  n ∈ Z.
   x = (2n + 1)π/4 or x = 2nπ ± π/3, where  n ∈ Z. 
    

Question-7 :-  Find the general solutions of sin 2x + cosx = 0.

  sin 2x + cosx = 0
  2 sin x cos x + cos x = 0
  cos x(2 sin x + 1) = 0
  cos x = 0 or 2 sin x + 1 = 0
  cos x = 0 or sin x = -1/2
  x = (2n + 1)π/2, where  n ∈ Z. 
  Now,
  sin x = -1/2 
  sin x = sin(π + π/6)
  sin x = sin 7π/6
      x = 7π/6
  x = nπ + (-1)ⁿ 7π/6, where  n ∈ Z.
  Therefore,x = (2n + 1)π/2 or x = nπ + (-1)ⁿ 7π/6, where  n ∈ Z.
     

Question-8 :-  Find the general solutions of sec² 2x = 1– tan 2x

  sec² 2x = 1 – tan 2x
  1 + tan² 2x - 1 + tan 2x = 0
  tan² 2x + tan 2x = 0
  tan 2x(tan 2x + 1) = 0
  tan 2x = 0 or tan 2x + 1 = 0 
  tan 2x = 0
      2x = 0
      2x = nπ + 0, where  n ∈ Z
       x = nπ/2, where  n ∈ Z
  Now,
  tan 2x + 1 = 0 
  tan 2x = -1
  tan 2x = -tan π/4
  tan 2x = tan(π - π/4)
  tan 2x = tan 3π/4
      2x = 3π/4
  2x = nπ + 3π/4, where  n ∈ Z
   x = nπ/2 + 3π/8, where  n ∈ Z
  Therefore, x = nπ/2 or x = nπ/2 + 3π/8, where  n ∈ Z
   

Question-9 :-  Find the general solutions of sin x + sin 3x + sin 5x = 0

  sin x + sin 3x + sin 5x = 0
  (sin x + sin 5x) + sin 3x = 0
  2 sin 6x/2 . cos (-4x)/2 + sin 3x = 0    [sin a + sin b = 2 sin (a + b)/2 . cos (a - b)/2]
  2 sin 3x . cos (-2x) + sin 3x = 0
  2 sin 3x . cos 2x + sin 3x = 0           [cos(-x) = cos x]
  sin 3x(2 cos 2x + 1) = 0
  sin 3x = 0 or 2 cos 2x + 1 = 0
  sin 3x = 0
      3x = 0
  3x = nπ + 0, where  n ∈ Z
   x = nπ/3, where  n ∈ Z
  Now,
  2 cos 2x + 1 = 0
  cos 2x = -1/2
  cos 2x = -cos π/3
  cos 2x = cos(π - π/3)
  cos 2x = cos 2π/3
      2x = 2π/3
  2x = 2nπ ± 2π/3, where  n ∈ Z.
   x = nπ ± π/3, where  n ∈ Z.  
  Therefore, x = nπ/3 or x = nπ ± π/3, where  n ∈ Z.