To solve the equation $\cos(2x) = \sin(x)$ by rewriting it as a quadratic equation in $\sin(x)$, let's proceed as follows:

Step 1: Rewrite $\cos(2x)$ in terms of $\sin(x)$

Using the double-angle identity: $\cos(2x) = 1 - 2\sin^2(x)$ we can substitute this into the equation: $1 - 2\sin^2(x) = \sin(x)$

Step 2: Rearrange the equation

Bring all terms to one side to form a quadratic equation: $2\sin^2(x) + \sin(x) - 1 = 0$

Step 3: Solve the quadratic equation

Let $u = \sin(x)$. Then the equation becomes: $2u^2 + u - 1 = 0$

This is a quadratic equation in $u$, which we can solve using the quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = 1$, and $c = -1$. Substituting these values gives: $u = \frac{-1 \pm \sqrt{(1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}$ $u = \frac{-1 \pm \sqrt{1 + 8}}{4}$ $u = \frac{-1 \pm 3}{4}$

This gives two solutions: $u = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad u = \frac{-4}{4} = -1$

So, $\sin(x) = \frac{1}{2}$ or $\sin(x) = -1$.

Step 4: Solve for $x$

1. Case 1: $\sin(x) = \frac{1}{2}$

   The values of $x$ in the interval $-\pi \leq x \leq \pi$ for which $\sin(x) = \frac{1}{2}$ are: $x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6}$

2. Case 2: $\sin(x) = -1$

   The value of $x$ in the interval $-\pi \leq x \leq \pi$ for which $\sin(x) = -1$ is: $x = -\frac{\pi}{2}$

Final Answer

The solutions to the equation $\cos(2x) = \sin(x)$ in the interval $-\pi \leq x \leq \pi$ are: $x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}, \quad \text{and} \quad x = -\frac{\pi}{2}$

Would you like further details on any of these steps?

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Here are five related questions to deepen understanding:

1. How does the double-angle identity help simplify trigonometric equations?
2. What are the steps to solve a quadratic equation, and when is the quadratic formula particularly useful?
3. How can we determine the specific interval solutions for trigonometric equations like this?
4. What other trigonometric identities could be used to solve similar equations?
5. How would the solution differ if the interval was $0 \leq x \leq 2\pi$ instead?

Tip: When solving trigonometric equations, always check if using identities can transform the equation into a simpler form, such as a quadratic, which is often easier to solve.