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 sin² x – cos x = 1/4

Now let us simplify,

sin² x – cos x = 1/4

1 – cos² x – cos x = 1/4 [since, sin² x = 1 – cos² x]

4 – 4 cos² x – 4 cos x = 1

4cos² x + 4cos x – 3 = 0

Suppose cos x be ‘k’

Therefore,

4k² + 4k – 3 = 0

4k² – 2k + 6k – 3 = 0

2k(2k – 1) + 3(2k – 1) = 0

(2k – 1) + (2k + 3) = 0

(2k – 1) = 0 or (2k + 3) = 0

k = 1/2 or k = -3/2

cos x = 1/2 or cos x = -3/2

Then, we shall consider only cos x = 1/2. cos x = -3/2 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) 2 cos² x – 5 cos x + 2 = 0

Now let us simplify,

2 cos² x – 5 cos x + 2 = 0

Suppose cos x be ‘k’

2k² – 5k + 2 = 0

2k² – 4k – k +2 = 0

2k(k – 2) - 1(k - 2) = 0

(k – 2) (2k – 1) = 0

k = 2 or k = 1/2

cos x = 2 or cos x = 1/2

Then, we shall consider only cos x = 1/2. cos x = 2 is not possible.

Therefore,

cos x = cos 60° = cos π/3

x = 2nπ ± π/3

∴ the general solution is

x = 2nπ ± π/3, where n ϵ Z.

(iii) Given as 2 sin² x + √3 cos x + 1 = 0

Now let us simplify,

2 sin² x + √3 cos x + 1 = 0

2 (1 – cos² x) + √3 cos x + 1 = 0 [since, sin² x = 1 – cos² x]

2 – 2 cos² x + √3 cos x + 1 = 0

2 cos² x – √3 cos x – 3 = 0

Suppose cos x be ‘k’

2k² – √3 k – 3 = 0

2k² - 2√3 k + √3 k – 3 = 0

2k(k – √3) + √3(k – √3) = 0

(2k + √3) (k – √3) = 0

k = √3 or k = -√3/2

cos x = √3 or cos x = -√3/2

Then, we shall consider only cos x = -√3/2. cos x = √3 is not possible.

Therefore,

cos x = -√3/2

cos x = cos 150° = cos 5π/6

Thus, x = 2nπ ± 5π/6, where n ϵ Z.