Let's solve the equation:

$7\sin(2x) - 1 = 3\sin(2x) + 2$

Step 1: Move all terms involving $\sin(2x)$ to one side.

Subtract $3\sin(2x)$ from both sides:

$7\sin(2x) - 3\sin(2x) - 1 = 2$

This simplifies to:

$4\sin(2x) - 1 = 2$

Step 2: Isolate $\sin(2x)$.

Add 1 to both sides:

$4\sin(2x) = 3$

Now, divide both sides by 4:

$\sin(2x) = \frac{3}{4}$

Step 3: Solve for $x$.

Take the inverse sine of both sides:

$2x = \sin^{-1}\left(\frac{3}{4}\right)$

Using a calculator, $\sin^{-1}\left(\frac{3}{4}\right) \approx 0.848$ radians.

So,

$2x = 0.848 \quad \text{or} \quad 2x = \pi - 0.848 = 2.293 \quad \text{(since sine is positive in both the first and second quadrants)}.$

Step 4: Solve for $x$.

Divide both values by 2:

$x = \frac{0.848}{2} \approx 0.424 \quad \text{or} \quad x = \frac{2.293}{2} \approx 1.146$

Step 5: Find other solutions.

Since sine has a period of $2\pi$, we can find additional solutions by adding multiples of $\pi$ to $x$.

So the general solutions are:

$x = 0.424 + n\pi \quad \text{or} \quad x = 1.146 + n\pi$

Where $n$ is any integer.

Step 6: Determine the solutions in the interval $0 < x < 2\pi$.

For $x = 0.424 + n\pi$:

• For $n = 0$, $x \approx 0.424$
• For $n = 1$, $x \approx 0.424 + \pi \approx 3.566$

For $x = 1.146 + n\pi$:

• For $n = 0$, $x \approx 1.146$
• For $n = 1$, $x \approx 1.146 + \pi \approx 4.287$

Thus, the solutions in the interval $0 < x < 2\pi$ are approximately:

$x \approx 0.424, \, 1.146, \, 3.566, \, 4.287$

Final Answer:

The solutions are $x \approx 0.424, 1.146, 3.566, 4.287$ in the interval $0 < x < 2\pi$.

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Would you like further details on any part of the process? Or do you have any follow-up questions?

Related questions:

1. How does the period of the sine function influence the number of solutions?
2. How can we check if these solutions are valid by substituting them back into the original equation?
3. How do you solve trigonometric equations involving multiple angles (e.g., $\sin(3x)$)?
4. What are the general methods for solving equations with trigonometric identities?
5. How can we use the unit circle to visualize solutions to trigonometric equations?

Tip:

When solving trigonometric equations, always remember to check for all possible solutions within the given interval by considering the periodic nature of trigonometric functions.