How do you solve #cos^2 xsin^2 x=sin x#; for #pi<x<=pi#?
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To solve the equation (\cos^2(x)  \sin^2(x) = \sin(x)) for ( \pi < x \leq \pi ), follow these steps:

Rewrite (\cos^2(x)  \sin^2(x) = \sin(x)) using the Pythagorean identity (\cos^2(x) = 1  \sin^2(x)):
(1  \sin^2(x)  \sin^2(x) = \sin(x))

Combine like terms:
(1  2\sin^2(x) = \sin(x))

Move all terms to one side of the equation to set it to zero:
(2\sin^2(x) + \sin(x)  1 = 0)

This equation is a quadratic equation in terms of (\sin(x)). To solve it, you can use the quadratic formula:
(\sin(x) = \frac{b \pm \sqrt{b^2  4ac}}{2a})
Where (a = 2), (b = 1), and (c = 1).

Substitute (a), (b), and (c) into the quadratic formula:
(\sin(x) = \frac{1 \pm \sqrt{1^2  4(2)(1)}}{2(2)})

Simplify under the square root:
(\sin(x) = \frac{1 \pm \sqrt{9}}{4})
(\sin(x) = \frac{1 \pm 3}{4})

This yields two possible solutions:
(a) \sin(x) = \frac{1 + 3}{4} = \frac{1}{2})
(b) \sin(x) = \frac{1  3}{4} = 1)

Now, you need to find the corresponding values of (x) for each solution.
a) For (\sin(x) = \frac{1}{2}), the solutions occur when (x = \frac{\pi}{6}) and (x = \frac{5\pi}{6}).
b) For (\sin(x) = 1), the solution occurs when (x = \frac{\pi}{2}).

Check each solution in the original equation to ensure they are valid within the given domain.
Therefore, the solutions to the equation (\cos^2(x)  \sin^2(x) = \sin(x)) for ( \pi < x \leq \pi ) are (x = \frac{\pi}{6}), (x = \frac{5\pi}{6}), and (x = \frac{\pi}{2}).
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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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