Solve the equation: 0. = sin x + cos 2x - 1.

Given:

• \[ \sin x + \cos 2x - 1 = 0 \]

Solution:

1. Use the double angle identity for cosine: \[ \cos 2x = 1 - 2\sin^2 x \] Substitute this into the equation: \[ \sin x + (1 - 2\sin^2 x) - 1 = 0 \]
2. Simplify the equation: \[ \sin x - 2\sin^2 x = 0 \]
3. Factor out $$sin x$$: \[ \sin x (1 - 2\sin x) = 0 \]
4. Set each factor to zero and solve for $$x$$:
   • Case 1: $$\( \sin x = 0 \)$$
   This gives \[ x = n\pi \], where $$n$$ is an integer.
   • Case 2: $$\( 1 - 2\sin x = 0 \)$$
   \[ 2\sin x = 1 \] \[ \sin x = \frac{1}{2} \] This gives \[ x = \frac{\pi}{6} + 2k\pi \] or \[ x = \frac{5\pi}{6} + 2k\pi \], where $$k$$ is an integer.
5. Combine the solutions: The general solutions are \[ x = n\pi \], \[ x = \frac{\pi}{6} + 2k\pi \], and \[ x = \frac{5\pi}{6} + 2k\pi \], where $$n$$ and $$k$$ are integers.

Answer:

• \[ x = n\pi \]
• \[ x = \frac{\pi}{6} + 2k\pi \]
• \[ x = \frac{5\pi}{6} + 2k\pi \]