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

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Answer 1 · 2015-09-11T17:01:27

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

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

How do you solve 2 sin^2 (theta) + 3 sin (theta) + 1 = 02sin2(θ)+3sin(θ)+1=02 sin^2 (theta) + 3 sin (theta) + 1 = 0 from [0,2pi]?