To solve the equation $\csc x = \cot x + 1$ in the interval $[0, 2\pi)$, let’s proceed step by step:

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Step 1: Rewrite in terms of sine and cosine

Recall that: $\csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}.$ Substitute these into the equation: $\frac{1}{\sin x} = \frac{\cos x}{\sin x} + 1.$

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Step 2: Eliminate the denominator

Multiply through by $\sin x$ (valid if $\sin x \neq 0$): $1 = \cos x + \sin x.$

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Step 3: Rearrange the equation

Rearrange to: $\sin x + \cos x = 1.$

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Step 4: Use a trigonometric identity

Use the identity: $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right),$ where $x + \frac{\pi}{4}$ is a phase-shifted angle. Thus, the equation becomes: $\sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 1.$

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Step 5: Solve for $\sin\left(x + \frac{\pi}{4}\right)$

Divide through by $\sqrt{2}$: $\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$

The solutions to $\sin u = \frac{\sqrt{2}}{2}$ are: $u = \frac{\pi}{4}, \frac{3\pi}{4} \quad \text{(in one period)}.$

So: $x + \frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4}.$

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Step 6: Solve for $x$

Subtract $\frac{\pi}{4}$ from both sides: $x = 0, \frac{2\pi}{4} = \frac{\pi}{2}.$

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Step 7: Check the interval

Both solutions $x = 0$ and $x = \frac{\pi}{2}$ lie in $[0, 2\pi)$.

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Final Answer:

The solutions are: $\boxed{0, \frac{\pi}{2}}$

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Would you like a detailed explanation of any step or additional verification?
Here are 5 related questions to deepen your understanding:

1. How does the identity $\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})$ work?
2. Why is it necessary to restrict $\sin x \neq 0$ when multiplying by $\sin x$?
3. What are other methods to solve $\sin x + \cos x = 1$?
4. How do you confirm solutions in trigonometric equations?
5. How do phase shifts like $x + \frac{\pi}{4}$ affect trigonometric graphs?

Tip: Always verify solutions by substituting them back into