# How do you find the exact value of #4sin^2theta-3sintheta=1# in the interval #0<=theta<2pi#?

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To find the exact value of (4\sin^2(\theta) - 3\sin(\theta) = 1) in the interval (0 \leq \theta < 2\pi), you can rewrite it as a quadratic equation in terms of (\sin(\theta)). Let (x = \sin(\theta)), then solve the resulting quadratic equation (4x^2 - 3x - 1 = 0). Once you find the values of (x), substitute them back into (\sin(\theta)) to find the corresponding values of (\theta) within the given interval.

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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.

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.

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.

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