Question

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

Answer 1

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

Explanation:

Let `t=sinθ` hence we have that

`2t^2+3t+1=0=>2t^2+2t+t+1=0=>2t(t+1)+(t+1)=0=>(t+1)(2t+1)=0=>t=-1 and t=-1/2`

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

Remember that θ belongs to `[0,2pi]` hence

we have that `sinθ=-1=>θ=(3pi)/2`

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