Gunnar Bittersmann: 0 = cos²x -sin²x

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Hello out there!

Noch ’ne Möglichkeit, zur Lösung zu kommen:

cos 2x = cos²x - sin²x = 0
    2x = ½π + kπ           (k ∈ ℤ)
     x = ¼π + ½kπ

See ya up the road,
Gunnar

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