To solve the equation $1 - \cos(x) = -\sin(x)$, we can proceed with the following steps:

Step 1: Rearrange the equation

The given equation is: $1 - \cos(x) = -\sin(x)$ Rearrange this to: $1 = \cos(x) - \sin(x)$

Step 2: Use trigonometric identity

We can try to express $\cos(x) - \sin(x)$ in a simpler form using a trigonometric identity. We use the identity for a linear combination of sine and cosine:

$\cos(x) - \sin(x) = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right)$ This identity is derived by recognizing that $\cos(x) - \sin(x)$ can be rewritten as a cosine function with a phase shift.

So, we can rewrite the equation as: $1 = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right)$

Step 3: Solve for $x$

Now, solve for $x$: $\cos\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ We know that: $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Thus, we can write: x + \frac{\pi}{4} = \pm \frac{\pi}{4} + 2k\pi \quad \text{(where k is any integer)}

Step 4: Solve for $x$

Now, solving for $x$: $x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{4} = -\frac{\pi}{4} + 2k\pi$ Subtracting $\frac{\pi}{4}$ from both sides: $x = 2k\pi \quad \text{or} \quad x = -\frac{\pi}{2} + 2k\pi$

Final Answer:

The general solution for $x$ is: $x = 2k\pi \quad \text{or} \quad x = -\frac{\pi}{2} + 2k\pi$ where $k$ is any integer.

Would you like to explore any specific part of this solution, or do you have any further questions?

Here are 5 related questions:

1. What is the derivation of the identity $\cos(x) - \sin(x) = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right)$?
2. How can you solve trigonometric equations involving sine and cosine functions more efficiently?
3. What is the geometric interpretation of the equation $1 - \cos(x) = -\sin(x)$?
4. How do you handle trigonometric equations involving multiple angles (like $\cos(2x)$)?
5. What are the periodic properties of sine and cosine functions, and how do they affect the solutions?

Tip: Always try to look for known trigonometric identities to simplify equations before attempting to solve them directly. This can often make the problem much easier to handle!