Задание:
tg (x-15°) ctg (x+15°)=1/3
Решение:
tg (x-15)*ctg (x+15)=1/3tg (x-15)=(tg (x) -tg (15) / (1+tg (x) tg (15) tg (x+15)=(tg (x)+tg (15) / (1-tg (x) tg (15) ctg (x+15)=(1-tg (x) tg (15) / (tg (x)+tg (15) tg (x-15)*ctg (x+15)=(tg (x) -tg (15) (1-tg (x) tg (15) /[ (tg (x)+tg (15) (1+tg (x) tg (15) ]=1/3tg (30)=2tg (15) / (1-tg^2 (15)=1/sqrt (3) пусть z=tg (15) тогда 2z/ (1-z^2)=1/sqrt (3) z1=2-sqrt (3) и z2=-2-sqrt (3) <0 z2 не лежит в первой четверти, значит угол не 15 градусов tg (x)=y (y- (2-sqrt (3) (1-y*(2-sqrt (3) /[ (y+2-sqrt (3) (1+y*(2-sqrt (3) ]=1/3 (y+2-sqrt (3) <>0 и (1+y*(2-sqrt (3) <>03 (y- (2-sqrt (3) (1-y*(2-sqrt (3)=[ (y+2-sqrt (3) (1+y*(2-sqrt (3) ]3 (y- (2-sqrt (3) (1-y*(2-sqrt (3) -[ (y+2-sqrt (3) (1+y*(2-sqrt (3) ]=04 (sqrt (3) -2) (y-1) ^2=0 => y=1tg (x)=1 => x=pi/4+pi*k
Знаете другой ответ?