a) To solve the equation 2-cos(2x) + 3sin(x) = 0, we can use trigonometric identities to simplify it.

Using the double angle identity for cosine, cos(2x) = 1 - 2sin^2(x), we can rewrite the equation as:

2 - (1 - 2sin^2(x)) + 3sin(x) = 0

Simplifying further, we get:

2 - 1 + 2sin^2(x) + 3sin(x) = 0
1 + 2sin^2(x) + 3sin(x) = 0
2sin^2(x) + 3sin(x) + 1 = 0

Now, let's substitute sin(x) = t, where -1 ≤ t ≤ 1:

2t^2 + 3t + 1 = 0

This is a quadratic equation which can be factored as:

(2t + 1)(t + 1) = 0

Setting each factor to 0 gives us:

2t + 1 = 0 => t = -1/2
t + 1 = 0 => t = -1

Now, substitute back t = sin(x):

sin(x) = -1/2 or sin(x) = -1

This gives us the solutions: x = π/6 + 2πn, x = 3π/2 + 2πn, for integer n.

b) To solve the equation 26sin(x)cos(x) - cos(4x) + 7 = 0, we can use trigonometric identities to simplify it.

Using the double angle identity for cosine, cos(4x) = 1 - 2sin^2(2x), we can rewrite the equation as:

26sin(x)cos(x) - (1 - 2sin^2(2x)) + 7 = 0

Simplifying further, we get:

26sin(x)cos(x) - 1 + 2sin^2(2x) + 7 = 0
26sin(x)cos(x) + 2sin^2(2x) + 6 = 0

Now, using the double angle identity for sin, sin(2x) = 2sin(x)cos(x), we can rewrite the equation as:

13sin(2x) + 2sin^2(2x) + 6 = 0
2sin^2(2x) + 13sin(2x) + 6 = 0

Now, let's substitute sin(2x) = t, where -1 ≤ t ≤ 1:

2t^2 + 13t + 6 = 0

This is a quadratic equation which can be factored as:

(2t + 1)(t + 6) = 0

Setting each factor to 0 gives us:

2t + 1 = 0 => t = -1/2
t + 6 = 0 => t = -6

Now, substitute back t = sin(2x):

sin(2x) = -1/2 or sin(2x) = -6

This gives us the solutions: x = π/6 + 2πn, x = 7π/6 + 2πn, for integer n.