Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) Given as 4 sin² x – 8 cos x + 1 = 0

Now let us simplify,

4 sin² x – 8 cos x + 1 = 0

4(1 – cos² x) – 8 cos x + 1 = 0 [since, sin² x = 1 – cos² x]

4 – 4 cos² x – 8 cos x + 1 = 0

4 cos² x + 8 cos x – 5 = 0

Suppose cos x be ‘k’

4k² + 8k – 5 = 0

4k² - 2k + 10k – 5 = 0

2k(2k – 1) + 5(2k – 1) = 0

(2k + 5) (2k – 1) = 0

k = -5/2 = -2.5 or k = 1/2

cos x = -2.5 or cos x = 1/2

Now, we shall consider only cos x = 1/2. cos x = -2.5 is not possible.

Therefore,

cos x = cos 60° = cos π/3

x = 2nπ ± π/3

∴ the general solution is

Thus, x = 2nπ ± π/3, where n ϵ Z.

(ii) tan² x + (1 – √3) tan x – √3 = 0

Now let us simplify,

tan² x + (1 – √3) tan x – √3 = 0

tan² x + tan x – √3 tan x – √3 = 0

tan x (tan x + 1) – √3 (tan x + 1) = 0

(tan x + 1) ( tan x – √3) = 0

tan x = -1 or tan x = √3

As, tan x ϵ (-∞ , ∞) therefore both values are valid and acceptable.

tan x = tan (-π/4) or tan x = tan (π/3)

x = mπ – π/4 or x = nπ + π/3

∴ the general solution is

x = mπ – π/4 or nπ + π/3, where m, n ϵ Z.

(iii) 3 cos² x – 2√3 sin x cos x – 3 sin² x = 0

Now let us simplify,

3 cos² x – 2√3 sin x cos x – 3 sin² x = 0

3 cos² x – 3√3 sin x cos x + √3 sin x cos x – 3 sin² x = 0

3 cos x (cos x – √3sin x) + √3 sin x (cos x – √3 sin x) = 0

√3 (cos x – √3 sin x) (√3 cos x + sin x) = 0

cos x – √3 sin x = 0 or sin x + √3 cos x = 0

cos x = √3 sin x or sin x = -√3 cos x

tan x = 1/√3 or tan x = -√3

As, tan x ϵ (-∞ , ∞) therefore both values are valid and acceptable.

tan x = tan (π/6) or tan x = tan (-π/3)

x = mπ + π/6 or x = nπ – π/3

∴ the general solution is

x = mπ + π/6 or nπ – π/3, where m, n ϵ Z.

(iv) Given as cos 4x = cos 2x

Now let us simplify,

cos 4x = cos 2x

4x = 2nπ ± 2x

Therefore,

4x = 2nπ + 2x [or] 4x = 2nπ – 2x

2x = 2nπ [or] 6x = 2nπ

x = nπ [or] x = nπ/3

∴ the general solution is

Thus, x = nπ [or] nπ/3, where n ϵ Z.