Whenever you have an equation where the product of some terms (2 or more) is zero, then that means that if any one of them is zero, it solves the equation. That's the same as when you solved quadratic equations by factoring.
So in this case you have two terms: (sinx+1) and (2sin^2x-3sinx-2). Either one of them can be zero. So you set each one to 0, which will give you two separate equations, and then you solve each equation.
So first do the easy one.
sin x + 1 = 0
sin x = -1
Where is the sin function equal to -1 in the interval [0, 2π)? There's only one place, at π. So π is one solution to the original equation.
Now, if you look at the choices you are given, only one of them has π as one of the solutions, choice b. That says to me that b is the answer, without even having to solve the second equation.
If you want to solve the second equation though, let's go through it.
2sin2 x - 3sin x - 2 = 0
The best way is to factor this, which you can do the same way you factor regular polynomials.
To make it easier, let's change sin x to y, so it looks like something more familiar to you:
2y2 -3y - 2 = 0
I won't go through all the possibilities, but this factors to:
(2y + 1)(y - 2) = 0
Put sin x back in for y so we can see what it's really supposed to be.
(2 sin x + 1)(sin x - 2) = 0
So again, either one of the terms 2sin x + 1 or sin x - 2 can be 0 for this equation to be true. Let's solve each of them separately.
2 sin x + 1 = 0
2 sin x = -1
sin x = -1/2
sin x is -1/2 at two points in the interval: 2π/3 and 5π/3. If you're not sure why this is true, go back to the textbook material about the sin function as the angle goes around the circle from 0 to 2π.
So two more solutions to the original equation are 2π/3 and 5π/3.
Now solve the last equation.
sin x - 2 = 0
sin x = 2
But the sin function can NEVER be 2. It can never be more than 1 as a matter of fact.
So there is no solution to this last equation.
So the solutions we've found to the original equation are π, 2π/3 and 5π/3. And that's correct for choice b.