From the identity #cos^2(x)+sin^2(x) = 1#, we can get #cos^2(x) = 1-sin^2(x)#. We will make this substitution, and then solve the resulting quadratic equation.
#3cos^2(x)-5sin(x) = 1#
#=> 3(1-sin^2(x))-5sin(x) = 1#
#=> 3-3sin^2(x)-5sin(x) = 1#
#=> 3sin^2(x)+5sin(x) - 2 = 0#
#=> (3sin(x) - 1)(sin(x)+2) = 0#
#=> 3sin(x)-1 = 0 or sin(x)+2 = 0#
#=> sin(x) = 1/3 or sin(x) = -2#
Assuming #x in RR#, then #-1 <= sin(x) <= 1#, so we can disregard the second possibility. Thus we have #sin(x) = 1/3#. To get the final general solution, though, we must note that #sin(x)# takes on the value of #1/3# in both the first and second quadrants, at #arcsin(1/3)# and #pi-arcsin(1/3)#. Account for multiples of #2pi#, we get our final solution:
#:. x = arcsin(1/3)+2pin ~~0.3398+2pin, n in ZZ#
or
#x = pi-arcsin(1/3)+2pin ~~ 2.8018+2pin, n in ZZ#
