Using the t-formulae find the range of x in the interval [0,2pi] for which 3sinx +2cosx>3

We are tasked with finding the range of \( x \) in the interval \( [0, 2\pi] \) for which the inequality \( 3\sin x + 2\cos x > 3 \) holds, using the **t-formulae**. Let's break it down step by step:

### Step 1: Rewrite the inequality using the t-formulae

The **t-formulae** involve the substitution:
\[
t = \tan \left(\frac{x}{2}\right)
\]
With this substitution, we can express \( \sin x \) and \( \cos x \) in terms of \( t \) as follows:
\[
\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}
\]
Substitute these into the inequality \( 3\sin x + 2\cos x > 3 \).

### Step 2: Apply the substitution to the inequality
Substituting the t-formulae expressions into the inequality:

\[
3 \cdot \frac{2t}{1 + t^2} + 2 \cdot \frac{1 - t^2}{1 + t^2} > 3
\]

Simplify the left-hand side:
\[
\frac{6t + 2(1 - t^2)}{1 + t^2} > 3
\]
This simplifies to:
\[
\frac{6t + 2 - 2t^2}{1 + t^2} > 3
\]

### Step 3: Eliminate the denominator
Multiply both sides by \( 1 + t^2 \) to get rid of the denominator:
\[
6t + 2 - 2t^2 > 3(1 + t^2)
\]
Expanding both sides:
\[
6t + 2 - 2t^2 > 3 + 3t^2
\]

### Step 4: Simplify the inequality
Bring all terms to one side:
\[
-2t^2 - 3t^2 + 6t + 2 - 3 > 0
\]
This simplifies to:
\[
-5t^2 + 6t - 1 > 0
\]

### Step 5: Solve the quadratic inequality
We now solve the quadratic inequality \( -5t^2 + 6t - 1 > 0 \). First, solve the corresponding quadratic equation:
\[
-5t^2 + 6t - 1 = 0
\]
Using the quadratic formula:
\[
t = \frac{-6 \pm \sqrt{6^2 - 4(-5)(-1)}}{2(-5)} = \frac{-6 \pm \sqrt{36 - 20}}{-10} = \frac{-6 \pm \sqrt{16}}{-10}
\]
\[
t = \frac{-6 \pm 4}{-10}
\]
So, the solutions for \( t \) are:
\[
t = \frac{-6 + 4}{-10} = \frac{-2}{-10} = \frac{1}{5}, \quad t = \frac{-6 - 4}{-10} = \frac{-10}{-10} = 1
\]

Thus, the quadratic equation has roots at \( t = \frac{1}{5} \) and \( t = 1 \).

### Step 6: Analyze the sign of the quadratic
The inequality \( -5t^2 + 6t - 1 > 0 \) is positive between the roots \( \frac{1}{5} \) and \( 1 \). Therefore, \( t \) must lie in the interval:
\[
\frac{1}{5} < t < 1
\]

### Step 7: Convert back to \( x \)
Recall that \( t = \tan \left( \frac{x}{2} \right) \), so we have:
\[
\frac{1}{5} < \tan \left( \frac{x}{2} \right) < 1
\]

Now solve for \( x \):
1. For \( \tan \left( \frac{x}{2} \right) = \frac{1}{5} \):
   \[
   \frac{x}{2} = \tan^{-1} \left( \frac{1}{5} \right) \quad \Rightarrow \quad x = 2 \tan^{-1} \left( \frac{1}{5} \right)
   \]
   Using a calculator, \( \tan^{-1} \left( \frac{1}{5} \right) \approx 0.197 \). So, \( x \approx 2 \times 0.197 = 0.394 \).

2. For \( \tan \left( \frac{x}{2} \right) = 1 \):
   \[
   \frac{x}{2} = \frac{\pi}{4} \quad \Rightarrow \quad x = 2 \times \frac{\pi}{4} = \frac{\pi}{2}
   \]

Thus, the range of \( x \) is approximately:
\[
0.394 < x < \frac{\pi}{2}
\]

### Final Answer:
The range of \( x \) for which \( 3\sin x + 2\cos x > 3 \) is:
\[
x \in (0.394, \frac{\pi}{2})
\]

Would you like further clarification on any of the steps?

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Here are 5 related questions:
1. How do we solve trigonometric inequalities involving more complex expressions?
2. What are the advantages of using the t-formulae in solving trigonometric equations?
3. How can we use graphical methods to solve trigonometric inequalities?
4. What are the general steps to convert between trigonometric expressions and t-formulae?
5. Can we extend this method to other intervals, like \( [-\pi, \pi] \)?

**Tip**: When solving quadratic inequalities, always consider the sign of the quadratic expression between its roots.

Using the t-formulae find the range of x in the interval [0,2pi] for which 3sinx + 2cosx > 3

Discover how to solve the trigonometric inequality 3sin(x) + 2cos(x) > 3 in the interval [0, 2pi] using the t-formulae. This detailed solution covers the use of t = tan(x/2), substitution, and solving quadratic inequalities. Suitable for Grades 11-12.

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