Find an equation for the tangent to the curve at p and the horizontal tangent to the curve at q. y = 5 + cot x - 2 csc x 0 1 2 3 0 2 4 x y upper qupper p left parenthesis startfraction pi over 2 endfraction comma 3 right parenthesis

Answer with explanation:

The given function in x and y is,

  y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

Slope of tangent at (π/2,3)

Equation of tangent passing through (π/2,3) can be obtained by

⇒There will be no Horizontal tangent from the point (π/2,3).

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step-by-step explanation:

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let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

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