How do you solve #1=8cos(2x +1) 3#?
First add 3 to both sides to get:
Next, divide both sides by 8 to get:
Subtract 1 from both sides to get:
Divide both sides by 2 to get:
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To solve (1 = 8\cos(2x + 1)  3):

Add 3 to both sides: (1 + 3 = 8\cos(2x + 1)) (4 = 8\cos(2x + 1))

Divide both sides by 8: (\frac{4}{8} = \cos(2x + 1)) (\frac{1}{2} = \cos(2x + 1))

To find (2x + 1), use the inverse cosine function (arc cosine): (2x + 1 = \cos^{1}\left(\frac{1}{2}\right))

Solve for (2x + 1): (2x + 1 = \frac{\pi}{3}) or (2x + 1 = \frac{5\pi}{3})

Solve for (x): For (2x + 1 = \frac{\pi}{3}): (2x = \frac{\pi}{3}  1) (x = \frac{\pi}{6}  \frac{1}{2})
For (2x + 1 = \frac{5\pi}{3}): (2x = \frac{5\pi}{3}  1) (x = \frac{5\pi}{6}  \frac{1}{2})
So, the solutions are (x = \frac{\pi}{6}  \frac{1}{2}) and (x = \frac{5\pi}{6}  \frac{1}{2}).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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