Question

$5 \arcctg { ^{2} {x} } - 3\pi \: arcctgx - 2\pi ^{2} = 0$
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Answer 1

$5(arctgx)^{2}-3\pi arctgx-2\pi^{2}=0\\\\arctgx=m,-\frac{\pi }{2} <m<\frac{\pi }{2}\\\\5m^{2}-3\pi m-2\pi^{2}=0\\\\D=(-3\pi)^{2}-4*5*(-2\pi ^{2})=9\pi ^{2}+40\pi ^{2}=49\pi ^{2}=(7\pi)^{2} \\\\m_{1}=\frac{3\pi-7\pi }{10}=-\frac{2\pi }{5}\\\\m_{2}=\frac{3\pi+7\pi }{10}=\pi-neyd\\\\arctgx=-\frac{2\pi }{5}$