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**Soroban** Hello, xlepher!

Hope I read your equation correctly . . .

We have: . ¼[3·cot²(½θ) - 1] .= .A

Multiply by 4: . 3·cot²(½θ) - 1 .= .4A

Add 1: . 3·cot²(½θ) .= .4A + 1

Divide by 3: . cot²(½θ) .= .(4A + 1)/3

. . . . . . . . . . . . . . . . . . . . . . . . . _________

Take square roots: . cot(½θ) .= .±√(4A + 1)/3

. . . . . . . . . . . . . . . . . . . . . . . . . - - . . . _________

Take inverse cotangent: . ½θ .= .arccot[±√(4A + 1)/3]

. . . . . . . . . . . . . . . . . . . . . . . _________

Multiply by 2: . θ .= .2·arccot[±√(4A + 1)/3]

Edit: Too fast for me, Jhevon!