Solve the following on the interval 0≤θ≤2π. sec⁡2θ+sec⁡θ=0

Answer 1

Hi,

sec⁡2θ+sec⁡θ=0 as you know secθ = 1/cosθ and

sec2θ = 1/cos2θ

now back to question sec⁡2θ+sec⁡θ=0 ====>

1/cos2θ + 1/cosθ = 0 both sides times cos2θ.cosθ =====>

cosθ + cos 2θ = 0 =====> cos2θ = -cosθ =====>

cos2θ = cos(∏ - θ ) you know when 2 cosines are equal you can make a decision that the relation between angles are 2θ = 2k∏±(∏ - θ)

k=0 ---> 2θ = ±(∏ - θ)====> 2θ = (∏ - θ) ---> 3θ = ∏ ----> θ = Π/3 Β AND

2θ = -(∏ - θ) --> 2θ = -∏ + θ ----> θ = - Π that the same answer with θ = Π

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k = 1 ---> 2θ = 2∏±(∏ - θ)====> 2θ = 2∏ +(∏ - θ)= 3Π -θ ====> 3θ = 3Π ====> θ = Π we had it

2θ = 2∏ -(∏ - θ) = 2Π -Π +θ=Π+θ ====> θ = Π SO does not need to choose k= 2,3,....

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As a result the answers are Π, Π/3

I hope it is useful,

Minoo