# How do you solve 2 sin^2 (theta) + 3 sin (theta) + 1 = 0 from [0,2pi]?

##### 1 Answer

The solutions are θ_1=(3pi)/2, θ_2=(11pi)/6, θ_3=(7pi)/6

#### Explanation:

Let t=sinθ hence we have that

$2 {t}^{2} + 3 t + 1 = 0 \implies 2 {t}^{2} + 2 t + t + 1 = 0 \implies 2 t \left(t + 1\right) + \left(t + 1\right) = 0 \implies \left(t + 1\right) \left(2 t + 1\right) = 0 \implies t = - 1 \mathmr{and} t = - \frac{1}{2}$

Hence we have that sinθ=-1 and sinθ=-1/2.Solving these we get

Remember that θ belongs to $\left[0 , 2 \pi\right]$ hence

we have that sinθ=-1=>θ=(3pi)/2

and sinθ=-1/2=>θ=(11pi)/6 and θ=(7pi)/6