Posted by Argelia on .
solve for all values in [0,2pie)
2cos^2x5sinx+1=0

precal 
Steve,
2(1sin^2)5sin+1=0
2sin^2 + 5sin  3 = 0
(2sin1)(sin+3) = 0
so, sinx = 1/2 or 3
3 is not allowed, so
sinx = 1/2: x = pi/6 or 5pi/6