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 - $1-2si{n}^2(x)$ + 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$x$ - $1-2si{n}^2(2x)$ + 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.