Question
sin(3x) + 2cos(2x) = 0
Answers
GPT-4o mini
To solve the equation \( \sin(3x) + 2\cos(2x) = 0 \), we can start by isolating one of the trigonometric functions:
\[
\sin(3x) = -2\cos(2x)
\]
Next, we can use the double angle identity for cosine, which states that \( \cos(2x) = 1 - 2\sin^2(x) \) or \( \cos(2x) = 2\cos^2(x) - 1 \). In this case, it might be more straightforward to use the first identity. However, to keep things simpler, let's solve \( \cos(2x) \) directly by making substitutions.
Instead, let's consider the identities we have:
We know \( \sin(3x) = 3\sin(x) - 4\sin^3(x) \) and \( \cos(2x) = 2\cos^2(x) - 1 \) or \( \cos(2x) = 1 - 2\sin^2(x) \).
To investigate solutions, let's first find specific angles or values of \( x \) that satisfy our equation. We can guess and check or graph these functions, but another effective way is to note the cyclic nature of the sine and cosine functions.
1. **Possible Solutions**: Use commonly known angles:
- When \( 3x = \frac{3\pi}{2} + k\pi \) for the sine function.
- When \( 2x = \frac{\pi}{2} + k\pi \) for the cosine function.
2. **Finding Specific Values**:
Let's first rewrite the equations as:
- \( \sin(3x) = 0 \)
- \( 2\cos(2x) = 0\)
From \( \sin(3x) = 0 \):
\[
3x = n\pi \implies x = \frac{n\pi}{3} \quad \text{for } n \in \mathbb{Z}
\]
From \( 2\cos(2x) = 0 \):
\[
\cos(2x) = 0 \implies 2x = \frac{\pi}{2} + k\pi \implies x = \frac{\pi}{4} + \frac{k\pi}{2} \quad \text{for } k \in \mathbb{Z}
\]
3. **Conversion between \( x \)**:
To make sure that both equations satisfy:
- Set \( x = \frac{n\pi}{3} \) into \( \frac{\pi}{4} + \frac{k\pi}{2} \).
Next, substitute and check the overlaps to find valid solutions with specific \( n \) and \( k \).
### Final Values
We derive the intersections of the \( x = \frac{n\pi}{3} \) and \( x = \frac{\pi}{4} + \frac{k\pi}{2} \).
The overall solutions will be:
\[
x = \frac{n\pi}{3} \quad \text{and/or} \quad x = \frac{\pi}{4} + \frac{k\pi}{2}
\]
Combine these to map points over the unit circle for overlaps and intersections creating an easily solvable approach for periodic functions and checking cosine phases against sine repetitions.
The final solution can be validated within numerical solutions. Use the range if required to constrain values. You can check valid \( n \) and \( k \) across typical angles, \( x = \frac{\pi}{12}, \frac{5\pi}{12} \), and so on for values in degrees too \( (15^\circ, 75^\circ) \).
Make sure to verify your results across the ranges set for sine and cosine functions.
This will yield all angles satisfying the original equation from both sides.
\[
\sin(3x) = -2\cos(2x)
\]
Next, we can use the double angle identity for cosine, which states that \( \cos(2x) = 1 - 2\sin^2(x) \) or \( \cos(2x) = 2\cos^2(x) - 1 \). In this case, it might be more straightforward to use the first identity. However, to keep things simpler, let's solve \( \cos(2x) \) directly by making substitutions.
Instead, let's consider the identities we have:
We know \( \sin(3x) = 3\sin(x) - 4\sin^3(x) \) and \( \cos(2x) = 2\cos^2(x) - 1 \) or \( \cos(2x) = 1 - 2\sin^2(x) \).
To investigate solutions, let's first find specific angles or values of \( x \) that satisfy our equation. We can guess and check or graph these functions, but another effective way is to note the cyclic nature of the sine and cosine functions.
1. **Possible Solutions**: Use commonly known angles:
- When \( 3x = \frac{3\pi}{2} + k\pi \) for the sine function.
- When \( 2x = \frac{\pi}{2} + k\pi \) for the cosine function.
2. **Finding Specific Values**:
Let's first rewrite the equations as:
- \( \sin(3x) = 0 \)
- \( 2\cos(2x) = 0\)
From \( \sin(3x) = 0 \):
\[
3x = n\pi \implies x = \frac{n\pi}{3} \quad \text{for } n \in \mathbb{Z}
\]
From \( 2\cos(2x) = 0 \):
\[
\cos(2x) = 0 \implies 2x = \frac{\pi}{2} + k\pi \implies x = \frac{\pi}{4} + \frac{k\pi}{2} \quad \text{for } k \in \mathbb{Z}
\]
3. **Conversion between \( x \)**:
To make sure that both equations satisfy:
- Set \( x = \frac{n\pi}{3} \) into \( \frac{\pi}{4} + \frac{k\pi}{2} \).
Next, substitute and check the overlaps to find valid solutions with specific \( n \) and \( k \).
### Final Values
We derive the intersections of the \( x = \frac{n\pi}{3} \) and \( x = \frac{\pi}{4} + \frac{k\pi}{2} \).
The overall solutions will be:
\[
x = \frac{n\pi}{3} \quad \text{and/or} \quad x = \frac{\pi}{4} + \frac{k\pi}{2}
\]
Combine these to map points over the unit circle for overlaps and intersections creating an easily solvable approach for periodic functions and checking cosine phases against sine repetitions.
The final solution can be validated within numerical solutions. Use the range if required to constrain values. You can check valid \( n \) and \( k \) across typical angles, \( x = \frac{\pi}{12}, \frac{5\pi}{12} \), and so on for values in degrees too \( (15^\circ, 75^\circ) \).
Make sure to verify your results across the ranges set for sine and cosine functions.
This will yield all angles satisfying the original equation from both sides.