How do you solve # 2cosx+1=0#?
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To solve (2\cos(x) + 1 = 0), follow these steps:

Subtract 1 from both sides of the equation: [ 2\cos(x) = 1 ]

Divide both sides by 2: [ \cos(x) = \frac{1}{2} ]

Determine the angles where the cosine function equals (\frac{1}{2}). These values can be found in the interval ( [0, 2\pi] ) where cosine is negative.

The solutions for (x) in this interval are ( \frac{2\pi}{3} ) and ( \frac{4\pi}{3} ).
So, the solutions to the equation (2\cos(x) + 1 = 0) are (x = \frac{2\pi}{3} + 2\pi n) and (x = \frac{4\pi}{3} + 2\pi n), where (n) is an integer.
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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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