Complex Trigonometric Equation? I am having difficulty solving cos (z) = sqrt 2 In class it was mentioned using the quadratic formula to solve for the inverse cos function but I am not sure how to do that.

Yazmin Sims

Yazmin Sims

Answered question

2022-10-30

Complex Trigonometric Equation?
I am having difficulty solving
cos ( z ) = 2
In class it was mentioned using the quadratic formula to solve for the inverse cos function but I am not sure how to do that.

Answer & Explanation

Finnegan Stone

Finnegan Stone

Beginner2022-10-31Added 11 answers

Notice, we have
cos z = 2
e i z + e i z 2 = 2
e 2 i z + 1 2 e i z = 2
e 2 i z + 1 = 2 2 e i z
( e i z ) 2 2 2 e i z + 1 = 0
Above is the quadratic equation in terms of e i z hence using quadratic formula
e i z = 2 2 ± ( 2 2 ) 2 4 ( 1 ) ( 1 ) 2 ( 1 )
e i z = 2 ± 1
i z = ln ( 2 ± 1 )
z = i ln ( 2 ± 1 )
Josiah Owens

Josiah Owens

Beginner2022-11-01Added 3 answers

You have
e i z + e i z = 2 2 .
so
e 2 i z + 1 = 2 2 e i z .
Put w = e i z ; this becomes
w 2 + 1 = 2 2 w .
This is a quadratic.

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