See the following graph to convince yourself that on the interval from [0, 2pi) this equation has 8 solutions.
The question is how to you find those solutions algebraically?
Start by factoring out the sin(2x) from each term to reach:
sin2x[2cos2x + 1] = 0
By the zero product property either sin2x = 0 OR 2cos2x+1 = 0.
To solve sin2x = 0 we start by solving for 2x.
2x = arc sin 0
The angles that have sin of 0 are 0 + kpi, where k is any integer.
That means x = kpi/2 , where k is any integer. Let k = 0, 1, 2, ... until you are outside the target interval.
So we have x = 0, pi/2, pi, 3pi/2.
Now let's work on 2cos2x +1 = 0.
We get cos2x = -1/2.
So 2x = arc cos -1/2.
The angles that have a cos of -1/2 are 2pi/3 + 2kpi and 4pi/3 + 2kpi, where k is any integer.
Once again let k = 0, 1, 2... until you are outside the target interval.
When 2x = 2pi/3 + 2kpi, x = pi/3 + kpi, or pi/3, 4pi/3 (k=0, 1)
When 2x = 4pi/3 + 2kpi, x= 2pi/3 + kpi, or 2pi/3, 5pi/3 (k=0,1)
There you go.