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Posted
On 3/5/2025 at 8:32 AM, DimaMazin said:

Can such problem be solved?

Area A =area of the green segment.

x=-cos/(1-2cos)

y=sin/(1-2cos)

sin/(1-2cos)=(p-6a)/2

20250305_141254.jpg

I don't think your equation is correct.

Posted
11 hours ago, Genady said:

I don't think your equation is correct.

Maybe you think sin and cos are sin(a) and cos(a)? No. sin is sin(p/2 +a) and cos is cos(p/2 +a)  as it showed on diagram.

Posted
3 hours ago, DimaMazin said:

y = cos(a)/(1+2sin(a))

cos(a)/(1+2sin(a))=(p-6a)/2

My starting equation is


\(\frac b 2 - \frac {\sin b \sin c} {2 \sin d} = \frac e 2 - \frac {\sin e} 2\)

where

\(b=2a\)
\(2c=\pi - (\frac {\pi} 2 + a)\) 
\(d = \pi - b - c\)
\(e = \frac {\pi} 2 - a\)

I don't know if it's the same as yours.

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