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

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to solve 2sin^2x + 3sinx = 2, let sinx = a

then the equation becomes,

2a^2 +3a = 2

2a^2 +3a-2 = 0 factorising the equation , we get,

2a^2 +4a-1a-2 = 0

2a(a+2) -1(a+2) = 0

(2a-1) (a+2) = 0 ,this shows that

either 2a-1 =0 or a+2 = 0,

either a = 1/2 or a= -2, but a = sinx,

thus sinx = 1/2 or sinx = -2

thus we get x= sin^-1(1/2) = pi/6, or x= sin^-1(-2)=undefined

Q :Find x, 2sin²x+ 3sin x = 2

A :

If sin x = y

then 2y²+3y=2

2y²+3y-2=0

(2y-1)(y+2)=0

so, either 2y-1=0 or y+2=0

either y=0.5 or y=-2

y ⇒ sin x,

so, sin x = 0.5 or sin x =-2

x = sin ⁻¹(0.5) or x = sin⁻¹(2)

**x = pi/6** , ⇒ sin⁻¹(2) is undefined

therefore the answer is **pi/6**