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How do you solve #1=8cos(2x +1) -3#?

Answer 1

Nolan Alexander

First add 3 to both sides to get:

#4 = 8 cos(2x+1)#

Next, divide both sides by 8 to get:

#1/2 = cos(2x+1)#

If #cos(2x+1) = 1/2# then #2x+1 = 2n pi +- pi/3# where #n# is any integer.

Subtract 1 from both sides to get:

#2x = 2n pi +- pi/3 - 1#

Divide both sides by 2 to get:

#x = n pi +- pi/6 - 1/2#

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Answer 2

Madelyn Adams

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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Answer from HIX Tutor

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.