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Could anyone help me solve the following equation for $F$:

$$P=\left|\sqrt{F^2+G^2+2FG\cos(2a)}\right|$$

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2 Answers 2

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You should have $$F_1= \frac{-2 G \cos[2 a] - \sqrt 2 \sqrt{G^2 + 2 P^2 + G^2 \cos[4 a]}}{2},$$ and $$F_2= \frac{-2 G \cos[2 a] + \sqrt 2 \sqrt{G^2 + 2 P^2 + G^2 \cos[4 a]}}{2},$$ because your equations can be manipulated as a simple quadratic one.

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Squaring both sides, $$ P^2 = F^2 + G^2 +2FG\cos{2a} \implies F^2 + (2G\cos{2a})F+G^2-P^2=0 $$

This is a quadratic in $F$.

Solving, you get, $$ F = -G\cos{2a} \pm \sqrt{P^2-(G\sin{2a})^2} $$

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