Roots of trigonometric equation In the following trigonometric equation 1 + α 2 cos 2 ⁡ ( n θ ) = 0The complex solutions are cos ⁡ ( n θ ) = ± i / αSo I thought that the correspondant angles were n θ = arccos ⁡ ( i / α ) + 2 k π n θ = 2 π − arccos ⁡ ( i / α ) + 2 k πand n θ = arccos ⁡ ( − i / α ) + 2 k π n θ = 2 π − arccos ⁡ ( − i / α ) + 2 k πas usual in trigonometric equations containing cosine. But instead the solutions should be n θ = arccos ⁡ ( i / α ) + k πand n θ = arccos ⁡ ( − i / α ) + k πWhy?

Question

Roots of trigonometric equation In the following trigonometric equation   1  +      α    2        cos    2    ⁡  (  n  θ  )  =  0The complex solutions are  cos  ⁡  (  n  θ  )  =  ±  i      /    αSo I thought that the correspondant angles were  n  θ  =  arccos  ⁡  (  i      /    α  )  +  2  k  π  n  θ  =  2  π  −  arccos  ⁡  (  i      /    α  )  +  2  k  πand  n  θ  =  arccos  ⁡  (  −  i      /    α  )  +  2  k  π  n  θ  =  2  π  −  arccos  ⁡  (  −  i      /    α  )  +  2  k  πas usual in trigonometric equations containing cosine. But instead the solutions should be  n  θ  =  arccos  ⁡  (  i      /    α  )  +  k  πand  n  θ  =  arccos  ⁡  (  −  i      /    α  )  +  k  πWhy?

Theresa Wade · Accepted Answer

Note that   arccos  ⁡  (  z  )  =  π  −  arccos  ⁡  (  −  z  ).So  arccos  ⁡  (  i      /    α  )  +  (  2  k  −  1  )  π  =  −  arccos  ⁡  (  −  i      /    α  )  +  2  k  πand  arccos  ⁡  (  −  i      /    α  )  +  (  2  k  −  1  )  π  =  −  arccos  ⁡  (  i      /    α  )  +  2  k  πThus these are just two different ways of writing the same set of solutions.

Roots of trigonometric equation In the following trigonometric equation 1 + alpha^2 cos^2 (n theta) = 0

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