# How do you solve #2cos^2x-2sin^2x=1# and find all solutions in the interval #0<=x<360#?

Where,

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To solve the equation (2\cos^2(x) - 2\sin^2(x) = 1) and find all solutions in the interval (0 \leq x < 360), follow these steps:

- Use the trigonometric identity (\cos^2(x) + \sin^2(x) = 1) to rewrite the equation in terms of either sine or cosine.
- Substitute the equivalent expression for either (\cos^2(x)) or (\sin^2(x)) in terms of the other trigonometric function.
- Solve the resulting quadratic equation for either (\cos(x)) or (\sin(x)) within the given interval.
- Once you have found the values of (\cos(x)) or (\sin(x)), use inverse trigonometric functions to find the corresponding values of (x) within the given interval.
- Check for extraneous solutions and ensure all solutions are within the specified interval.

Following these steps, you'll find the solutions for the equation in 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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