# How do you find the points of intersection of #r=3(1+sintheta), r=3(1-sintheta)#?

Hence the points of interaction are:

# (3, sin npi) #

If we consider the interval

# (3, 0) # and# (3, pi) #

We have:

# r = 3(1+sintheta) #

# r = 3(1-sintheta) #

At any point of intersection both equations are simultaneously satisfied, so we have:

# 3(1+sintheta) = 3(1-sintheta) #

# :. 1+sintheta = 1-sintheta #

# :. 2sintheta = 0 #

# :. sintheta = 0 #

# :. theta = npi #

And with

# r = 3(1+0) #

# :. r =3 #

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To find the points of intersection of the polar curves ( r = 3(1 + \sin(\theta)) ) and ( r = 3(1 - \sin(\theta)) ), set the two equations equal to each other and solve for ( \theta ). Once you find the values of ( \theta ), plug them back into either equation to find the corresponding values of ( r ). These pairs of ( r ) and ( \theta ) will represent the points of intersection.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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