solve sin(x) = cos(2x) for -2pi is less than or equal to x is lessthan or equal to 0 
give your answer in the form of a comma separated list express numbers in exact form use symbolic notaion and fractions where needed make sure to note Rewrite the equation using the double‑angle formula for the cosine function.

cos(2𝑥)=1−2sin2(𝑥)
 
Then express the equation as a combination of two equations. Solve them on the same interval and combine the obtained solutions.

To solve the equation \(\sin(x) = \cos(2x)\) in the interval \([-2\pi, 0]\), we follow the steps below:

### Step 1: Rewrite the equation using the double-angle formula for \(\cos(2x)\)
The double-angle identity for cosine is:
\[
\cos(2x) = 1 - 2\sin^2(x)
\]
Substitute this into the original equation:
\[
\sin(x) = 1 - 2\sin^2(x)
\]

### Step 2: Rearrange the equation
Move 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 \( u = \sin(x) \). The equation becomes:
\[
2u^2 + u - 1 = 0
\]
This is a quadratic equation. We can solve it using the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 2\), \(b = 1\), and \(c = -1\). Substituting these values, we get:
\[
u = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}
\]
\[
u = \frac{-1 \pm \sqrt{1 + 8}}{4}
\]
\[
u = \frac{-1 \pm \sqrt{9}}{4}
\]
\[
u = \frac{-1 \pm 3}{4}
\]
Thus, the solutions for \(u\) are:
\[
u = \frac{-1 + 3}{4} = \frac{1}{2}, \quad u = \frac{-1 - 3}{4} = -1
\]

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

### Step 4: Solve for \(x\)
#### Case 1: \(\sin(x) = \frac{1}{2}\)
The solutions to \(\sin(x) = \frac{1}{2}\) in the interval \([-2\pi, 0]\) are:
\[
x = -\frac{\pi}{6}, \quad x = -\frac{5\pi}{6}
\]

#### Case 2: \(\sin(x) = -1\)
The solution to \(\sin(x) = -1\) in the interval \([-2\pi, 0]\) is:
\[
x = -\frac{3\pi}{2}
\]

### Step 5: Combine the solutions
The solutions to \(\sin(x) = \cos(2x)\) in the interval \([-2\pi, 0]\) are:
\[
x = -\frac{\pi}{6}, -\frac{5\pi}{6}, -\frac{3\pi}{2}
\]

Thus, the solution set is:
\[
x = -\frac{\pi}{6}, -\frac{5\pi}{6}, -\frac{3\pi}{2}
\]

Would you like to go over the steps in more detail or have any questions?

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Here are 5 related questions you might find interesting:
1. How would the solution change if the interval was \([0, 2\pi]\)?
2. Can you derive the general solution for \(\sin(x) = \cos(2x)\) for all \(x\)?
3. How does the solution method change if we deal with different trigonometric identities?
4. What is the geometric interpretation of the solutions on the unit circle?
5. How do other double-angle formulas (like for \(\sin(2x)\)) affect solving trigonometric equations?

**Tip**: For solving trigonometric equations, always look for opportunities to use identities like the double-angle or Pythagorean identities to simplify the expressions.

Solve sin(x) = cos(2x) for -2π ≤ x ≤ 0. Rewrite the equation using the double-angle formula for the cosine function.

Explore the step-by-step solution for the trigonometric equation sin(x) = cos(2x) over the interval -2π ≤ x ≤ 0. Learn how to use the double-angle formula for cosine and solve the resulting quadratic equation. This problem is ideal for students in Grades 10-12, covering key trigonometric concepts and identities.

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