solve cos^2x - sin^2x = sin x; 0 less than equal to x less than 2pi

To solve the equation \(\cos^2 x - \sin^2 x = \sin x\) in the interval \(0 \leq x < 2\pi\), we follow these steps:

1. Use the trigonometric identity:
   \[
   \cos^2 x - \sin^2 x = \cos(2x)
   \]
   Substituting this into the equation gives:
   \[
   \cos(2x) = \sin x
   \]

2. We need to solve:
   \[
   \cos(2x) = \sin x
   \]

3. Let \(y = 2x\), so the equation becomes:
   \[
   \cos y = \sin\left(\frac{y}{2}\right)
   \]

4. Recall the identity for \(\sin\left(\frac{y}{2}\right)\):
   \[
   \sin\left(\frac{y}{2}\right) = \pm \sqrt{\frac{1 - \cos y}{2}}
   \]

5. Substitute this into the equation:
   \[
   \cos y = \pm \sqrt{\frac{1 - \cos y}{2}}
   \]

6. Square both sides to eliminate the square root:
   \[
   \cos^2 y = \frac{1 - \cos y}{2}
   \]

7. Multiply both sides by 2:
   \[
   2 \cos^2 y = 1 - \cos y
   \]

8. Rearrange to form a quadratic equation in terms of \(\cos y\):
   \[
   2 \cos^2 y + \cos y - 1 = 0
   \]

9. Solve the quadratic equation using the quadratic formula, \(ax^2 + bx + c = 0\):
   \[
   \cos y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \(a = 2\), \(b = 1\), and \(c = -1\):
   \[
   \cos y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}
   \]
   \[
   \cos y = \frac{-1 \pm \sqrt{1 + 8}}{4}
   \]
   \[
   \cos y = \frac{-1 \pm 3}{4}
   \]
   So we get:
   \[
   \cos y = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \cos y = \frac{-4}{4} = -1
   \]

10. Now, solving for \(y\) (or \(2x\)) in each case:
    - For \(\cos y = \frac{1}{2}\):
      \[
      y = 2k\pi \pm \frac{\pi}{3}, \quad k \in \mathbb{Z}
      \]

    - For \(\cos y = -1\):
      \[
      y = (2k+1)\pi, \quad k \in \mathbb{Z}
      \]

11. Convert back to \(x\) by dividing \(y\) by 2:
    - For \(\cos y = \frac{1}{2}\):
      \[
      2x = 2k\pi \pm \frac{\pi}{3} \implies x = k\pi \pm \frac{\pi}{6}
      \]

    - For \(\cos y = -1\):
      \[
      2x = (2k+1)\pi \implies x = \frac{(2k+1)\pi}{2}
      \]

12. Determine all \(x\) in the interval \(0 \leq x < 2\pi\):
    - From \(x = k\pi \pm \frac{\pi}{6}\):
      \[
      x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}
      \]

    - From \(x = \frac{(2k+1)\pi}{2}\):
      \[
      x = \frac{\pi}{2}, \frac{3\pi}{2}
      \]

Combining all solutions, we have:
\[
x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{2}, \frac{3\pi}{2}
\]

Would you like more details or have any questions?

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**Tip:** Always double-check your solutions by substituting them back into the original equation to ensure they satisfy the given conditions.

Learn how to solve the trigonometric equation cos^2x - sin^2x = sin x step-by-step. This problem involves trigonometric identities and quadratic equations. Suitable for advanced high school students and college-level trigonometry courses.

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