# Problem (1) For what value of #x#, #sinx+cosx=2# Problem (2) For what value of #x#, #sinxcosx=1#?

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Problem (1)

Problem (2)

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Problem (1): For what value of x, (\sin(x) + \cos(x) = 2)?

The equation (\sin(x) + \cos(x) = 2) has no solution because the maximum value of (\sin(x)) is 1 and the maximum value of (\cos(x)) is also 1, so the maximum possible sum of (\sin(x) + \cos(x)) is (1 + 1 = 2), but this only occurs when both (\sin(x)) and (\cos(x)) are at their maximum values, which never happens simultaneously since their maximum values occur at different points in the unit circle ((\sin(x)) at (x = \frac{\pi}{2}) and (\cos(x)) at (x = 0)).

Problem (2): For what value of x, (\sin(x)\cos(x) = 1)?

This equation also has no real solution because the product (\sin(x)\cos(x)) has a maximum value of (\frac{1}{2}). This is derived from the identity (\sin(2x) = 2\sin(x)\cos(x)), implying that (2\sin(x)\cos(x) = \sin(2x)). The maximum value of (\sin(2x)) is 1, so the maximum of (2\sin(x)\cos(x)) is 1, making the maximum of (\sin(x)\cos(x) = \frac{1}{2}), occurring when (x = \frac{\pi}{4}) or (x = \frac{3\pi}{4}), not 1. Therefore, there's no real value of (x) that satisfies (\sin(x)\cos(x) = 1).

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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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