Question

Решить уравнение: tgx-ctg((7pi)/6)=tg((7pi)/8) -√3×tg(pi/8)×tgx

Answer 1

Сначала найдем известные числа
ctg(7pi/6) = ctg(pi/6) = √3
$sin( \frac{7\pi}{8} ) = sin(\pi - \frac{\pi}{8} ) = sin( \frac{\pi}{8} )= \sqrt{ \frac{1-cos(\pi/4)}{2} } = \sqrt{ \frac{1- \sqrt{2}/2 }{2} } = \frac{ \sqrt{2- \sqrt{2} } }{2}$
$cos( \frac{7\pi}{8} )=cos(\pi- \frac{\pi}{8} )=-cos( \frac{\pi}{8} )=-\sqrt{ \frac{1+cos(\pi/4)}{2} } = -\sqrt{ \frac{1+ \sqrt{2}/2 }{2} } =$
$= -\frac{ \sqrt{2+ \sqrt{2} } }{2}$

$tg( \frac{7\pi}{8} )= \frac{sin( \frac{7\pi}{8} )}{cos( \frac{7\pi}{8} )} = -\frac{ \sqrt{2- \sqrt{2} } }{2}:\frac{ \sqrt{2+ \sqrt{2} } }{2}= -\sqrt{ \frac{2- \sqrt{2}}{2+ \sqrt{2}} } = -\sqrt{ \frac{(2- \sqrt{2})^2}{4-2} } =$
$=-\frac{2- \sqrt{2}}{ \sqrt{2} }=1- \sqrt{2}$

$tg( \frac{\pi}{8} )= \frac{sin( \frac{\pi}{8} )}{cos( \frac{\pi}{8} )} = \frac{ \sqrt{2- \sqrt{2} } }{2}:\frac{ \sqrt{2+ \sqrt{2} } }{2}= \sqrt{ \frac{2- \sqrt{2}}{2+ \sqrt{2}} } = \sqrt{ \frac{(2- \sqrt{2})^2}{4-2} } =$
$=\frac{2- \sqrt{2}}{ \sqrt{2} }=\sqrt{2} -1$

Подставляем
$tgx- \sqrt{3}=1- \sqrt{2} - \sqrt{3}*( \sqrt{2} -1) *tgx$
$tgx+ \sqrt{3}( \sqrt{2} -1)*tgx=1- \sqrt{2}+ \sqrt{3}$
$tgx*(1+ \sqrt{6} - \sqrt{3} )=1- \sqrt{2}- \sqrt{3}$
$tgx= \frac{1- \sqrt{2}- \sqrt{3}}{1+ \sqrt{6} - \sqrt{3}} = \frac{(1- \sqrt{2}- \sqrt{3})(1+ \sqrt{6} + \sqrt{3})}{(1+ \sqrt{6})^2 - 3}=$
$=\frac{1-\sqrt{2}- \sqrt{3}+ \sqrt{6}-2 \sqrt{3}-3 \sqrt{2}+ \sqrt{3}- \sqrt{6} -3 }{1+2 \sqrt{6}+6-3} = \frac{-2-4 \sqrt{2}-2 \sqrt{3}}{4+2 \sqrt{6} } =- \frac{1+2 \sqrt{2}- \sqrt{3} }{2+ \sqrt{6} }=$
$=- \frac{(1+2 \sqrt{2}- \sqrt{3})(2- \sqrt{6} ) }{4-6}= \frac{2+4 \sqrt{2}-2 \sqrt{3}- \sqrt{6} -4 \sqrt{3}+3 \sqrt{2}}{2} = \frac{2+7 \sqrt{2} -6 \sqrt{3} - \sqrt{6} }{2}$
$x=arctg(\frac{2+7 \sqrt{2} -6 \sqrt{3} - \sqrt{6} }{2})+ \pi *k$
Если я ничего не напутал. Сложное решение получилось.