Multiply both sides by #sin(x)#

#2sin^2(x) + 2 = 5sin(x)#

Replace #sin(x)# for #y#

#2y^2 + 2 = 5y#

Isolate #y#

#2y^2 - 5y + 2 = 0#

Solve the quadratic

#y = (5+-sqrt(25-4*2*2))/4 = (5+-sqrt(25-16))/4#

#y = (5+-3)/4#

So we have that #y = 2# or #y = 1/2#

Since we're dealing with trig functions, we know that #-1 < y<1#, so the only valid answer is #y = 1/2# or

#sin(x) = 1/2#

We know from the special angles / unit circle that there are 2 angles in which this true.

#30º# or #pi/6# and #150º# or #(5pi)/6#

However, any of these angles plus or minus a full loop would give the same answer, so we say that set of solutions #S# is

#S = {pi/6 + 2pik, (5pi)/6 + 2pik}#

or, in degrees,

#S = {30º + 360kº, 150 + 360kº}#

Where #k# is an integer constant. It's also important to remember that since we were dividing by a sine at first, we could make a not of the restriction that we can't have any value of #x# such that #sin(x) = 0#, so #x != pik# or, in degrees #x != 180kº# where #k# is an integer constant