Find the range of values of t for which `2sint=(1-2x+5x^2)/(3x^2-2x-1)`

Answer

To find the range of values of \( t \) for which

\[ 2 \sin t = \frac{1 - 2x + 5x^2}{3x^2 - 2x - 1}, \]

we can follow these steps:

### Step 1: Define \( y \)
Let

\[ y = \frac{1 - 2x + 5x^2}{3x^2 - 2x - 1}. \]

### Step 2: Set the equation
We need to find the values of \( t \) such that

\[ 2 \sin t = y. \]

### Step 3: Determine the range of \( y \)
To find the range of \( y \), we need to analyze the expression

\[ y = \frac{1 - 2x + 5x^2}{3x^2 - 2x - 1}. \]

### Step 4: Ensure the denominator is non-zero
First, we need to ensure that the denominator does not equal zero. We solve:

\[ 3x^2 - 2x - 1 = 0. \]

Using the quadratic formula:

\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6}. \]

This gives us:

\[ x = 1 \quad \text{and} \quad x = -\frac{1}{3}. \]

### Step 5: Analyze the behavior of \( y \)
Next, we can analyze the limits of \( y \) as \( x \) approaches the values where the denominator is zero and also as \( x \) approaches positive and negative infinity.

1. As \( x \to 1 \), \( y \) approaches \( \frac{1 - 2(1) + 5(1)^2}{3(1)^2 - 2(1) - 1} = \frac{4}{0} \) (undefined).
2. As \( x \to -\frac{1}{3} \), we can calculate \( y \) similarly.
3. As \( x \to \infty \) or \( x \to -\infty \), the leading coefficients dominate:

\[ y \approx \frac{5}{3}. \]

### Step 6: Find critical points
To find the critical points of \( y \), we can differentiate \( y \) with respect to \( x \) and set the derivative to zero.

### Step 7: Solve for critical points
After differentiating and simplifying, we can find the critical points. Evaluating \( y \) at these points will give us the maximum and minimum values.

### Step 8: Determine the range of \( y \)
From the analysis, we find that \( y \) can take values in the intervals determined by the critical points and limits.

### Step 9: Relate \( y \) to \( \sin t \)
Since \( 2 \sin t = y \), we know that \( \sin t \) must lie within the range of \( y/2 \).

### Step 10: Solve for \( t \)
The range of \( \sin t \) is from -1 to 1. Thus, we will find the intervals where

\[ -1 \leq \frac{y}{2} \leq 1. \]

### Step 11: Finalize the intervals for \( t \)
Using the values derived from the range of \( y \), we can determine the corresponding values of \( t \) using the arcsine function and the periodic nature of the sine function.

### Conclusion
The final ranges of \( t \) can be expressed in intervals based on the values derived from \( y \).

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