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 sin 2x = √3/2

Now, let us simplify,

sin 2x = √3/2

= sin(π/3)

∴ the general solution is

2x = nπ + (-1)ⁿ π/3, where n ϵ Z.

Thus, x = nπ/2 + (-1)ⁿ π/6, where n ϵ Z.

(ii) cos 3x = 1/2

Now, let us simplify,

cos 3x = 1/2

= cos (π/3)

∴ the general solution is

3x = 2nπ ± π/3, where n ϵ Z.

x = 2nπ/3 ± π/9, where n ϵ Z.

(iii) sin 9x = sin x

Now, let us simplify,

Sin 9x – sin x = 0

By using transformation formula,

Sin A – sin B = 2 cos(A + B)/2 sin(A - B)/2

Therefore,

= 2 cos(9x + x)/2 sin(9x - x)/2

=> cos 5x sin 4x = 0

Cos 5x = 0 or sin 4x = 0

Then, let us verify both the expressions,

Cos 5x = 0

Cos 5x = cos π/2

5x = (2n + 1)π/2

x = (2n + 1)π/10, where n ϵ Z.

sin 4x = 0

sin 4x = sin 0

4x = nπ

x = nπ/4, where n ϵ Z.

∴ the general solution is

x = (2n + 1)π/10 or nπ/4, where n ϵ Z.

(iv) sin 2x = cos 3x

Now, let us simplify,

sin 2x = cos 3x

cos (π/2 – 2x) = cos 3x [since, sin A = cos (π/2 – A)]

π/2 – 2x = 2nπ ± 3x

π/2 – 2x = 2nπ + 3x [or] π/2 – 2x = 2nπ – 3x

5x = π/2 + 2nπ [or] x = 2nπ – π/2

5x = π/2 (1 + 4n) [or] x = π/2 (4n – 1)

x = π/10 (1 + 4n) [or] x = π/2 (4n – 1)

∴ the general solution is

x = π/10 (4n + 1) [or] x = π/2 (4n – 1), where n ϵ Z.

(v) tan x + cot 2x = 0

Now, let us simplify,

tan x = – cot 2x

tan x = – tan (π/2 – 2x) [since, cot A = tan (π/2 – A)]

tan x = tan (2x – π/2) [since, – tan A = tan -A]

x = nπ + 2x – π/2

x = nπ – π/2

∴ the general solution is

x = nπ – π/2, where n ϵ Z.

(vi) tan 3x = cot x

Now, let us simplify,

tan 3x = cot x

tan 3x = tan(π/2 – x) [since, cot A = tan(π/2 – A)]

3x = nπ + π/2 – x

4x = nπ + π/2

x = nπ/4 + π/8

∴ the general solution is

Thus, x = nπ/4 + π/8, where n ϵ Z.