How do you solve #cos^2 x-sin^2 x=sin x#; for #-pi<x<=pi#?

To solve the equation (\cos^2(x) - \sin^2(x) = \sin(x)) for ( -\pi < x \leq \pi ), follow these steps:

1. 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))

2. Combine like terms:

   (1 - 2\sin^2(x) = \sin(x))

3. Move all terms to one side of the equation to set it to zero:

   (2\sin^2(x) + \sin(x) - 1 = 0)

4. 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).

5. Substitute (a), (b), and (c) into the quadratic formula:

   (\sin(x) = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)})

6. Simplify under the square root:

   (\sin(x) = \frac{-1 \pm \sqrt{9}}{4})

   (\sin(x) = \frac{-1 \pm 3}{4})

7. This yields two possible solutions:

   (a) \sin(x) = \frac{-1 + 3}{4} = \frac{1}{2})

   (b) \sin(x) = \frac{-1 - 3}{4} = -1)

8. 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}).

9. 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}).