(cos x5)(cos x+1) = 0 what values in the interval (0, 2 pi) make this true?
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To find the values of ( x ) in the interval ( (0, 2\pi) ) that make ( (\cos x  5)(\cos x + 1) = 0 ) true, we solve each factor separately for ( \cos x ) and then find the corresponding values of ( x ).
First, set each factor equal to zero:

( \cos x  5 = 0 ) ( \Rightarrow \cos x = 5 ) (which is not possible since the cosine function's range is [1, 1]).

( \cos x + 1 = 0 ) ( \Rightarrow \cos x = 1 ) ( \Rightarrow x = \pi ) (since ( \cos(\pi) = 1 )).
So, the only solution in the interval ( (0, 2\pi) ) is ( x = \pi ).
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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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