Question 961977
For[0,2π)
2sin^2x+7cosx-5=0
2(1-cos^2x)+7cosx-5=0
2-2cos^2x+7cosx-5=0
2cos^2x-7cosx+3=0
(2cosx-1)(cosx-3)
2cosx-1=0
cosx=1/2
x=π/3, 5π/3
or 
cosx-3=0
cpsx=3  (reject,( -1 ≤ cosx ≤ 1))