Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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

1 Answer

Konstantinos Michailidis

Sep 11, 2015

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#

Related questions

See all questions in Solving Trigonometric Equations

Impact of this question

43488 views around the world

You can reuse this answer
Creative Commons License