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
