Предмет: Алгебра, автор: Lizyn4ik

Написать уравнение касательной к графику функции y= 2tgx/2 в точке 7п/3

Ответы

Автор ответа: Аноним

y(7π/3)=2tg(7π/6)=2tgπ/6=2/√3
y`=1/cos²x/2
y`(7π/3)=1/cos²7π/6=1/cos²π/6=1:3/4=4/3
Y=2√3/3+4/3(x-7π/3)

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