How do you solve #2cosx+3=0# in the interval #0<=x<=2pi#?
No solution
Hence, there is no solution.
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To solve the equation (2\cos(x) + 3 = 0) in the interval (0 \leq x \leq 2\pi), follow these steps:

Subtract 3 from both sides of the equation: [ 2\cos(x) = 3 ]

Divide both sides by 2: [ \cos(x) = \frac{3}{2} ]

Since cosine is negative in the second and third quadrants, and the interval is (0 \leq x \leq 2\pi), we need to find the solutions in those quadrants.

Find the reference angle for ( \cos(x) = \frac{3}{2} ). Since the cosine function is not defined for values greater than 1 or less than 1, there are no real solutions in this interval. Therefore, the equation has no solution in the given interval.
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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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