Y=-ctg(x+π/3) M(0;-√3)

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вопрос

To find the equation of the tangent line to the curve y=-ctg(x+π/3) at the point M(0, -√3), we need to find the derivative of the function y=-ctg(x+π/3) and then evaluate it at x=0 to find the slope of the tangent line.

Find the derivative of y=-ctg(x+π/3):
y = -ctg(x+π/3)
Take the derivative with respect to x:
y' = -c * (-csc^2(x+π/3))

Evaluate the derivative at x=0:
When x=0, csc(π/3) = 2/√3
y' = -c (-2/√3)^2
y' = -c 4/3
y' = -4c/3

Find the slope of the tangent line at M(0, -√3):
The tangent line's slope is equal to the derivative evaluated at x=0:
m = -4c/3

The equation of the tangent line:
m = -4c/3
Given point M(0, -√3), substitute into the point-slope form equation y-y1 = m(x-x1):
y-(-√3) = (-4c/3)(x-0)
y+√3 = (-4c/3)x
y = (-4c/3)x - √3

Therefore, the equation of the tangent line to the curve y=-ctg(x+π/3) at the point M(0, -√3) is y = (-4c/3)x - √3.