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