Finding the Equation of a Tangent: y=2sinx at P(5pie/6,1)

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by the function y=2sin(x) at the point P(5π/6, 1). The correct equation of the tangent line is derived using the gradient obtained from the derivative of the function. The textbook solution provided is √3x + y - 1 - 5√3π/6 = 0, confirming that both the gradient and y-intercept are essential components in defining the line.

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Ry122
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How do i find the equation of the tangent to the graph given by the equation at the point
P(5pie/6),1) y=2sinx
The answer in the back of the textbook says.
sqroot3X + y - 1 - 5sqroot3pie/6 = 0
 
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what do you need to define a line? gradient and intercept...
 
Gradient can be given by the derivative, so find the derivative. And the y-intercept is quite straight forward.
 

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