a) Let's rewrite the equation as a quadratic equation in terms of cosx:

4cos^2 x + 4cosx - 3 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula:

The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 4, b = 4, and c = -3.

Plugging those values into the formula, we get:

cosx = (-4 ± √(4^2 - 44(-3))) / 2*4
cosx = (-4 ± √(16 + 48)) / 8
cosx = (-4 ± √64) / 8
cosx = (-4 ± 8) / 8

This gives us two possible solutions:

cosx = (-4 + 8) / 8 = 4 / 8 = 0.5cosx = (-4 - 8) / 8 = -12 / 8 = -1.5 (which is not a valid value for cosx)

Therefore, the solution to the equation 4cos^2 x = 3 - 4cosx is cosx = 0.5.

b) Let's rewrite the equation in terms of sinx:

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

2 - 2sin^2 x + 3cosx - 3 = 0

-2sin^2 x + 3cos x - 1 = 0

Now we can use the identity sin^2 x + cos^2 x = 1 to replace sin^2 x:

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

-2 + 2cos^2 x + 3cos x - 1 = 0

2cos^2 x + 3cos x - 3 = 0

Now we have a quadratic equation in terms of cos x. We can solve this equation similarly to part a), which would give us the solutions for sin x.