# `2sin^2 (x) + 3sin(x) + 1 = 0` Find all the solutions of the equation in the interval `[0,2pi)`.

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mathace | Certified Educator

Solve the equation `2sin^2(x)+3sin(x)+1=0` in the interval [0,2pi).

`2sin^2(x)+3sin(x)+1=0`

`(2sin(x)+1)(sin(x)+1)=0`

Set each factor equal to zero and solve for the x value(s).

`2sin(x)+1=0`

`2sin(x)=-1`

`sin(x)=-1/2`

`x=(7pi)/6,x=(11pi)/6`

`sin(x)+1=0`

`sin(x)=-1`

`x=(3pi)/2`