2Sin^2(x)+3sin(x)=2solve

2Sin^2(x)+3sin(x)=2solve

2Sin^2(x)+3sin(x)=2solve

2Sin^2(x)+3sin(x)=2solve
to solve 2sin^2x + 3sinx = 2,   let sinx = athen the equation becomes,2a^2 +3a = 22a^2 +3a-2 = 0   factorising the equation , we get,2a^2 +4a-1a-2 = 02a(a+2) -1(a+2) = 0(2a-1) (a+2) = 0 ,this shows thateither 2a-1 =0 or a+2 = 0,either a = 1/2  or a= -2, but a = sinx,thus sinx = 1/2 or sinx = -2thus we get x= sin^-1(1/2) = pi/6,   or x= sin^-1(-2)=undefined