Solve -\sin^2\theta + 2\cos\theta +\cos^2\theta = 0 Using the quadratic formula. This

Caitlyn Cole

Caitlyn Cole

Answered question

2022-04-22

Solve
sin2θ+2cosθ+cos2θ=0
Using the quadratic formula.
This is what you should get
θ=cos1(1+32)

Answer & Explanation

Yaretzi Odom

Yaretzi Odom

Beginner2022-04-23Added 16 answers

Substitute sin2θ=1cos2θ to obtain
(1cos2θ)+2cosθ+cos2θ=0
2cos2θ+2cosθ1=0
Substitute t=cosθ. This is a quadratic equation
2t2+2t1=0
with solutions
t=12(1±3)
Note however that t=cosθ[1,1], that is t1/2(-1-3)=-1.366[-1,1] So the solutions are
cosθ=12(31)
θ=±arccos(12(31))+2πn nZ

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