Question 886036
Solve the equation on the interval [0,2pi)
2sin^2x=-3sinx+5, in terms of pi and use integers and/or fractions 
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{{{2sin^2(x)=-3sin(x)+5}}}
{{{2sin^2(x)+3sin(x)-5=0}}}
{{{(2sin(x)+5)(sin(x)-1)=0}}}
2sin(x)+5=0
sin(x)=-5/2(reject,(-1 < sin(x) < 1))
or
sin(x)-1=0
sin(x)=1
x=&#960;/2